Optical Properties of Solids 2nd Ed by Mark Fox

Name: Tapas Banerji

Email: [email protected]

IP: 129.82.195.78

Optical Properties of Solids Second Edition

Mark Fox Oxford University Press, 2010 SOLUTIONS TO EXERCISES These notes contain detailed solutions to the Exercises at the end of 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 solutions 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 2010 °

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

Name: Tapas Banerji


IP: 129.82.195.78

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 1.29 to obtain R = 0.041. The transmission is calculated from eqn 1.9 with R = 0.041 to obtain T = 92%. (1.2) From Table 1.4 we read that the refractive indices of fused silica and dense flint glass are 1.46 and 1.746 respectively. The reflectivities are then calculated from eqn 1.29 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 1.25 and 1.26 to convert ²˜r 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 × 107 m s−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 interface, and 90% of the light is absorbed when exp(−αl) = 0.1 at the operating wavelength. With α = 1.3 × 105 m−1 at 850 nm, we then find l = 1.8 × 10−5 m = 18 µm. (1.5) We are given n = 3.68 and we can use eqn 1.19 to work out κ = αλ/4π = 0.083. We then use eqn 1.29 to find R = 0.328. Since αl = 2.6, we do not need to consider multiple reflections and we can just use eqn 1.8 to find the transmission. This gives: T = (1 − 0.328)2 exp(−1.3 × 2) = 0.034. The optical density is calculated from eqn 1.11 as 0.434 × 1.3 × 2 = 1.1. (1.6) 99.8% absorption in 10 m means exp(−αl) = 0.002, and hence α = 0.62 m−1 . We use eqn 1.19 to find κ = αλ/4π = 3.5 × 10−8 . We thus have n ˜ = 1.33 + i 3.5 × 10−8 . The real and imaginary parts of ²˜r are found from eqns 1.23 and 1.24 respectively, and we thus obtain ²˜r = 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) (a) In the incoherent limit, we just add the intensities of the beams. The intensities of the beams transmitted after multiple reflections are shown

1


Name: Tapas Banerji


IP: 129.82.195.78

incident light R1 I0

R2 e-al

I0(1-R1)

I0(1-R1)e-a l

I0R1

I0(1-R1)R2e

I0(1-R1)R2e-a l I0(1-R1) R2R1e-3a l

I0(1-R1)2R2e-2a l

I0(1-R1)2 R22R1 e-4a l

transmitted light

-2a l

I0(1-R1) R2R1e-2a l I0(1-R1) R22R1e-4a l

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

I0(1-R1) R22R1e-3a l I0(1-R1) R22R12e-5a l

I0(1-R1) R22R12e-4a l

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

I0(1-R1) R23R12e-5a l

reflected light

I0(1-R1) (1-R2) R22R12 e-5a l Figure 1: Multiple reflections in the incoherent limit, as considered in Exercise 1.8. in Fig. 1. The transmitted intensity is given by: It

= I0 (1 − R1 )(1 − R2 )e−αl + I0 (1 − R1 )(1 − R2 )R1 R2 e−3αl + I0 (1 − R1 )(1 − R2 )R12 R22 e−5αl + · · · ¡ ¢ = I0 (1 − R1 )(1 − R2 )e−αl 1 + R1 R2 e−2αl + (R1 R2 )2 e−4αl + · · · , ∞ X = I0 (1 − R1 )(1 − R2 )e−αl (R1 R2 e−2αl )k , k=0

= I0 (1 − R1 )(1 − R2 )e where we used the identity transmissivity is thus: T =

−αl

1 , 1 − R1 R2 e−2αl

P∞ k=0

xk = 1/(1 − x) in the last line. The

It (1 − R1 )(1 − R2 )e−αl = . I0 1 − R1 R2 e−2αl

(b) We have an air-medium-air situation, and so it will be the case that R1 = R2 ≡ R. We need to compare the exact formulae given in eqns 1.6 and 1.9 with the approximate one that neglects multiple reflections given in eqn 1.8. (i) With α = 0 the extinction coefficient κ will also be zero. We then calculate the reflectivity from eqn 1.29 to be: R=

2.42 (3.4 − 1)2 = = 0.30 . (3.4 + 1)2 4.42 2


Name: Tapas Banerji


IP: 129.82.195.78

Hence the exact transmission from eqn 1.9 is given by: Texact =

1 − 0.30 0.70 = = 54% . 1 + 0.30 1.30

The approximate result with multiple reflections neglected is found from eqn 1.8 with α = 0: Tapprox = (1 − R)2 = (1 − 0.30)2 = (0.70)2 = 49% . Thus by neglecting multiple reflections, the transmission is underestimated by a factor of about 10%. (ii) If we assume n À κ, we can ignore the terms in κ in the formula for the reflectivity (eqn 1.29) and so just obtain R = 0.30 as in part (a). The exact transmission is calculated from eqn 1.6 to be: Texact =

(1 − 0.30)2 e−1 = 18.2% . 1 − 0.302 e−2

The formula with multiple reflections neglected (eqn 1.8) gives: Tapprox = (1 − 0.30)2 e−1 = 18.0% . The transmission is thus underestimated by a factor of 1%. (iii) Following the same method as in part (b)(i), we find: R Texact Tapprox

= 0.772 /2.772 = 0.077 , = (1 − 0.077)/(1 + 0.077) = 85.7% , = (1 − 0.077)2 = 85.2% .

The error is thus −0.6%. (c) It is intuitively obvious that it is valid to neglect multiple reflections when the absorption is strong (i.e. αl & 1), since the reflected beams will be very weak. Part (b)(ii) confirms this. For the case of transparent materials (i.e. α = 0), it will be valid to neglect multiple reflections when R is small, since then it will be valid to approximate 1/(1 + R) as (1 − R), and eqn 1.8 with α = 0 will be equivalent to eqn 1.9. The reflectivity will be small when (n − 1) is small (see eqn 1.29), and so we conclude that multiple reflections are insignificant for small (n − 1). This point is illustrated by comparing parts (b)(i) and (iii). (1.9) In contrast to the previous exercise, we must now add up the electric field amplitudes of the beams and consider their relative phases. Let r and t be the amplitude reflection and transmission coefficients for going from air to the medium, and r0 and t0 be the equivalent quantities for going from the medium to air. If the absorption coefficient is α, the amplitude will decay by a factor of e−αl/2 during a single pass of the medium, since I ∝ E 2 . For simplicity we define x = e−αl/2 eiΦ/2 , where Φ = 4πnl/λ is the round-trip phase shift, so that the complex amplitude changes by a 3


Name: Tapas Banerji


t

r incident amplitude

IP: 129.82.195.78

t'

r'

x = eif/2 e-al / 2

E0 t E0 tx E0

r E0 tx2r' E0

txr' E0

tx2r'2E0

tx3r'2E0

tx4r'3 E0

tx3r'3E0

tx4r'4 E0

tx5r'4 E0

transmitted amplitude tt'x E0

tt'x2r'E0

tt'x4r'3E0

reflected amplitude

tx5r'5 E0

tt'x3r'2E0

tt'x5r'4 E0

Figure 2: Multiple reflections in the coherent limit, as considered in Exercise 1.9. factor x after one pass through the medium. (a) With the definitions above, the transmitted amplitude E t is given by (see Fig. 2): Et

= = =

tt0 xE 0 + tt0 x3 r02 E 0 + tt0 x5 r04 E 0 + · · · , tt0 xE 0 (1 + x2 r02 + x4 r04 + · · · ) , ∞ X tt0 xE 0 (x2 r02 )k , k=0

=

tt0 x E0 , 1 − x2 r02

P∞ where we used k=0 ak = 1/(1 − a) in the last line. The transmitted intensity is then given by: I t ∝ E t (E t )∗ = |E 0 |2

(1 −

|x|2 |tt0 |2 . − x2 r02 )∗

x2 r02 )(1

Now x2 = e−αl eiΦ , r = −r0 (due to the π phase shift on going from a more dense to a less dense medium), tt0 = 1 − r2 , and r2 = r02 = R. Therefore

4


Name: Tapas Banerji


IP: 129.82.195.78

we have: T

= = = =

It |E t |2 = , I0 |E 0 |2 (1 − R)2 e−αl , −αl (1 − Re eiΦ )(1 − Re−αl e−iΦ ) (1 − R)2 e−αl , −αl (1 − Re (eiΦ + e−iΦ ) + R2 e−2αl ) (1 − R)2 e−αl , (1 − 2Re−αl cos Φ + R2 e−2αl )

where we used eiϕ + e−iϕ = 2 cos ϕ in the last line. (b) For the reflectivity, we proceed as in part (a). From Fig. 2 it is apparent that the reflected amplitude E r is given by: Er

= rE 0 + tt0 x2 r0 E 0 + tt0 x4 r03 E 0 + · · · , = E 0 (r + r0 tt0 x2 [1 + x2 r02 + x4 r04 + · · · ]) , µ ¶ r0 tt0 x2 = E0 r + , 1 − x2 r02 P∞ where we again used k=0 ak = 1/(1 − a). On inserting r0 = −r and tt0 = 1 − r2 = 1 − R, we find: µ ¶ µ ¶ x2 (1 − R) 1 − x2 Er = E0 r 1 − = E r . 0 1 − x2 R 1 − x2 R Therefore, with x2 = e−αl eiΦ , and again using eiϕ + e−iϕ = 2 cos ϕ, we find: reflectivity

|E r |2 , |E 0 |2 (1 − e−αl eiΦ )(1 − e−αl e−iΦ ) = R , (1 − Re−αl eiΦ )(1 − Re−αl e−iΦ ) (1 − 2e−αl cos Φ + e−2αl ) = R . (1 − 2Re−αl cos Φ + R2 e−2αl )

=

(c) In the limit with α = 0, the reflectivity and transmission formulae just reduce to the standard ones for a Fabry–Perot etalon (see e.g. Hecht chapter 9): (1 − R)2 , limα→0 (T ) = (1 − 2R cos Φ + R2 ) limα→0 (reflectivity) =

2R(1 − cos Φ) . (1 − 2R cos Φ + R2 )

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

5


Name: Tapas Banerji


IP: 129.82.195.78

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

(1 − R)2 2R(1 − cos Φ) + I0 (1 − 2R cos Φ + R2 ) (1 − 2R cos Φ + R2 ) (1 − R)2 + 2R(1 − cos Φ) I0 . (1 − 2R cos Φ + R2 ) (1 − 2R + R2 + 2R − 2R cos Φ) I0 , (1 − 2R cos Φ + R2 ) (1 + R2 − 2R cos Φ) , I0 (1 − 2R cos Φ + R2 ) I0 .

= I0 = = = =

(d) If αl À 1, then e−αl ¿ 1. Since R ≤ 1, we can neglect the second and third terms in the denominator, and so the transmission is just: T = (1 − R)2 e−αl , as in eqn 1.8. (e) For negligible absorption, we expect to observe thin-film fringes. The transmission will be as given in part (c): limα→0 (T ) =

(1 − R)2 . (1 − 2R cos Φ + R2 )

When Φ = 2mπ, where m is an integer, we have constructive interference, with: (1 − R)2 (1 − R)2 T = = = 1. (1 − 2R + R2 ) (1 − R)2 Since Φ = 4πnl/λ, this constructive interference condition occurs when 2nl = mλ, that is, the round trip path length is equal to an integer number of wavelengths. The other extreme occurs when Φ = (2m + 1)π: T =

(1 − R)2 (1 − R)2 = . (1 + 2R + R2 ) (1 + R)2

At this point, we have 2nl = (m + 1/2)λ0 , so that the waves from each round trip are out of phase and interfere destructively. Therefore, as the wavelength is scanned, the transmission oscillates between unity and (1 − R)2 /(1 + R)2 , as in a Fabry–Perot etalon. (1.10) This problem is meant to be a rough attempt to model the optical properties of GaAs near its band edge. The band gap is taken to be 1.42 eV, as appropriate for GaAs at room temperature, and the form of the absorption is meant to be rough fit to the data in Fig. 3.9, with C chosen to give the correct absorption around 2 eV, namely ∼ 4 × 106 m−1 . The reflectivity and transmissivity have to be calculated from the two formulae given in parts (a) and (b) of the previous exercise. In principle, for wavelengths below the band edge, we have to work out the extinction coefficient from the absorption via eqn 1.19, and then work out the reflection 6



IP: 129.82.195.78

1.0

Transmission or Reflectivity

Name: Tapas Banerji

0.8 Transmission Reflectivity

0.6 0.4 0.2 0.0 600

700

800

900

1000

Wavelength (nm)

Figure 3: Transmission and reflectivity of a 2µm thick platelet, as considered in Exercise 1.10. coefficient from eqn 1.29 with both the terms in n and κ included. However, the maximum value of κ in this exercise is only 0.19, which occurs at 600 nm, where α = 4 × 106 m−1 , and (0.19)2 is negligible compared to (3.5 − 1)2 . Therefore, in fact we can just work out R from the real part of the refractive index and use R = 2.52 /4.52 = 0.31 at all wavelengths. The transmissivity and reflectivity calculated in this way are plotted in Fig. 3. For wavelengths below the band edge (i.e. 870–1000 nm), the medium is transparent, and we observe Fabry–Perot fringes. Peaks in the transmission with T = 1 occur at wavelengths that satisfy 2nl = mλ, (m = integer), i.e. at 875 nm (m = 16), 933 nm (m = 15), and 1000 nm (m = 14). The reflectivity at these wavelengths is equal to zero. Peaks in the reflectivity and minima in the transmission are observed in between the transmission maxima. The minimum value of the transmission is (1 − R)2 /(1 + R)2 = 0.28, implying that the maximum value of the reflectivity is 0.72. For wavelengths below the band edge, the absorption increases rapidly, and the interference fringes are rapidly damped, with multiple reflections becoming negligible. The reflectivity settles to the value determined by the front surface (i.e. 31%), while the transmission decreases exponentially according to eqn 1.8. The transmissivity at 600 nm where α = 4×106 m−1 is (0.69)2 exp(−8) = 1.6 × 10−4 . (1.11) In the incoherent limit, the transmissivity of a transparent plate is given by eqn 1.9 as (1 − R)/(1 + R), while the reflectivity is obtained from

7


Name: Tapas Banerji


IP: 129.82.195.78

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

(n − 1)2 (n + 1)2 − (n − 1)2 4n = = . (n + 1)2 (n + 1)2 (n + 1)2

Similarly: 1+R=1+

(n − 1)2 (n + 1)2 + (n − 1)2 2(n2 + 1) = = . 2 2 (n + 1) (n + 1) (n + 1)2

We therefore have: T =

1−R 4n (n + 1)2 2n = × = 2 . 2 1+R (n + 1) 2(n2 + 1) (n + 1)

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

(n1 − n2 )2 . (n1 + n2 )2

Note that this reduces to R = (n − 1)2 /(n + 1)2 for the case of an airmedium interface with n1 = 1 and n2 = n. (cf. eqn 1.29 with κ = 0.) For the three cases considered here we have: • air → film: n1 = 1, n2 = 2.5, so R = (1.5/3.5)2 = 18%. • film → substrate: n1 = 2.5, n2 = 1.5, so R = (2.5 − 1.5)2 /(2.5 + 1.5)2 = (1/4)2 = 6%. • substrate → air: n1 = 1.5, n2 = 1, so R = (0.5/2.5)2 = 4%. (1.13) For a thick sample it is appropriate to use eqn 1.8 for the transmission. We take the log10 of eqn 1.8 to obtain (using log10 x = loge x/ loge 10): − log10 (T ) = −2 log10 (1 − R) + αl/ loge 10 . We can then substitute αl/ loge 10 from eqn 1.11 to obtain the required result. If the medium is transparent at λ0 then we will have that Tλ0 = (1 − R)/(1 + R) for the incoherent limit, 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, just above the absorption edge we are trying to measure. With this assumption, we can work out the value of R from a measurement of the transmission at λ0 , and then use this value in the formula derived in the exercise to work out the optical density from a measurement of the transmission at λ. Measurements of T (λ) and T (λ0 ) thus allow the optical density to be determined. The absorption coefficient can then be determined from the optical density by using eqn 1.11. (1.14) With σ = 6.6 × 107 Ω−1 m, and ω = 1.88 × 1013 rad/s, we find ²˜r = ²r + i 3.97 × 105 . The imaginary part of ²˜r is very large, and hence the

8


Name: Tapas Banerji


IP: 129.82.195.78

approximation ²2 À √ √ ²1 is a good one. In this approximation, we have (with i = (1 + i)/ 2): p √ n ˜ = ²˜r = (3.97 × 105 )1/2 i = 445(1 + i) . By inserting n = κ = 445 into eqn 1.29, we then obtain R = 99.6 %. (1.15) We use the same formula for the complex dielectric constant as in the previous exercise. With ω = 1.88 × 1013 rad/s, we find n ˜ 2 = ²˜r = ²1 + i 2.94 × 105 . Since ²2 À ²1 , this implies: p √ n ˜ = ²˜r = (2.94 × 105 )1/2 i = 383(1 + i) . We then find from eqn 1.19 that α = 4πκ/λ = 4.8 × 107 m−1 . Beer’s law means that we set exp(−αl) = 0.5 for a drop in intensity by a factor of 2, giving l = 1.4 × 10−8 m = 14 nm. (1.16) It is apparent from eqn 1.29 that R = 1 when n = 1 and κ = 0. For zero reflectivity we thus require ²˜r = (n + iκ)2 = 1. (1.17) (a) We convert wavelengths to photon energies using E = hc/λ to obtain the energy level scheme shown in Fig. 4. It is thus apparent that 0.294 eV of energy is dissipated during each absorption / emission process. (b) When the quantum efficiency is 100%, every absorbed photon produces 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. (c) For a luminescent quantum efficiency of 50% the number of photons emitted drops by a factor of 2 compared to part (b), and so the light power emitted falls to 4 W. The remaining 6 W of the absorbed power is dissipated as heat.

relaxation (0.294 eV) absorption 1.459 eV

emission 1.165 eV

Figure 4: Energy level scheme for Exercise 1.17. (1.18) 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 hν out = hν in − hν phonon . With ν = c/λ, we then find λout = 521 nm. 9


Name: Tapas Banerji


IP: 129.82.195.78

(1.19) The transmission is given by eqn 1.12, with the wavelength dependence of the scattering cross section given by eqn 1.13. At 850 nm we have 10% transmission, so that N σs l = 2.30. Since σs ∝ λ−4 , the scattering crosssection is 11.1 times larger at 850 nm than at 1550 nm, and so we have N σs l = 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. (1.20) We again use eqn 1.12 to calculate the transmission, setting exp(−N σs l) = 0.5. This gives N σs l = 0.69, which implies l = 3.5 m for the given values of N and σs . If the wavelength is reduced by a factor of two, Rayleigh’s scattering law (eqn 1.13) implies that σs increases by a factor of 16. The length required for the same transmission is thus smaller by a factor of 16: i.e. l = 3.5/16 = 0.22 m. (1.21) Birefringence is an example of optical anisotropy as discussed in Section 1.5.1, and also in Section 2.5. Ice is a uniaxial crystal, and therefore has preferential axes, making optical anisotropy possible. Water, by contrast, is a liquid and has no preferential axes. The optical properties must therefore be isotropic, making birefringence impossible.

10


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 2 Classical propagation (2.1) We envisage two displaced masses as shown in Fig. 5. 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 m1

d2 x1 = −Ks (x1 − x2 ) dt2

and

d2 x2 = −Ks (x2 − x1 ) . dt2 Divide the equations by m1 and m2 respectively and subtract them to obtain: µ ¶ 1 1 d2 (x − x ) = −K + (x1 − x2 ) . 1 2 s dt2 m1 m2 m2

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

d2 x = −Ks x . dt2

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

m1

rest

x1

displaced

m2

x2

Figure 5: 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 F0 e−iωt , and look for solutions of the form x(t) = x0 e−iωt . On substituting into the equation of motion we then obtain: m(−ω 2 − iωγ + ω02 )x0 e−iωt = F0 e−iωt , which implies: x(t) =

F0 1 e−iωt . m (ω02 − ω 2 − iωγ) 11


Name: Tapas Banerji


IP: 129.82.195.78

The phase factor comes from the factor of [ω02 − ω 2 − iωγ]−1 . On multiplying the top and bottom by the complex conjugate, we find: ω02

1 (ω 2 − ω 2 ) + iωγ = 20 2 2 . 2 − ω − iωγ (ω0 − ω ) + (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: ωγ tan θ = 2 . (ω0 − ω 2 ) This implies that the displacement of the oscillator is of the form: x(t) ∼ eiθ e−iωt = e−i(ωt−θ) , which shows that the oscillator has a relative phase lag of: θ = tan−1 [ωγ/(ω02 − ω 2 )] . (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 × 1015 rad/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. On using the value of N given in the exercise, 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 ) N e2 1 = ×f, 2n 2n²0 m0 γω0

where f = 9 × 10−5 is the oscillator strength. For the absorption line we 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 find κ(ω0 ) = 8.6 × 10−6 . We finally obtain the absorption at the line centre (405 nm) from eqn 1.19 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). 12


Name: Tapas Banerji


IP: 129.82.195.78

(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 2 see from eqn 2.16 that ²2 (ω0 ) = N e /²0 m0 γω0 , so that the absorption is (cf. eqn 1.19): α(ω0 ) =

4πκ(ω0 ) ²2 (ω0 ) N e2 = 4π × ÷ (2πc/ω0 ) = . λ 2n n²0 m0 γc

This shows that it is the linewidth that determines the peak absorption strength per oscillator. The oscillator strength is, of course, also important. (2.6) The data can be analysed by comparison with Fig. 2.4 with the assumption that the oscillator strength is unity. √ (a) The low frequency refractive index corresponds to ²st . With n = 2.43 for ω ¿ ω0 from the data, we find ²st = 5.9. (b) The resonant frequency is the mid point of the “wiggle”, i.e. 5.0 × 1012 Hz. (c) The natural frequency is given by eqn 2.2, which implies Ks = µω02 . The reduced mass µ is given by: 1/µ = 1/m1 + 1/m2 = 1/23 + 1/35.5 amu−1 , which gives µ = 14 amu = 2.33 × 10−26 kg. With ω0 = 2πν0 = 3.1 × 1013 rad/s, we find Ks = 23 kg s−2 . The restoring force is given by F = −Ks x, which implies |F | = 23 N for x equal to unity. (d) The oscillator density can be found from eqn 2.19. ²st = 5.9 has been found in part (a), and ²∞ can be read from the graph for ω À ω0 as ²∞ = n2 = (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 × 1028 m−3 . (e) γ is equal to the shift between the maximum and minimum in the refractive index in angular frequency units. We can only make a rough estimate of γ because the data does not follow a simple line shape. The damping rate depends strongly on the frequency, which is why the resonance line is asymmetric. By comparison with Fig. 2.4 we find ∆ν ∼ 1×1012 Hz, and hence γ = 2π∆ν ∼ 6 × 1012 s−1 . (f) The result of Exercise 2.5 tells us that α = N e2 /n²0 µγc 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 × 106 m−1 . (2.7) We start by re-writing eqn 2.25 as: 1 dk = . vg dω

13


Name: Tapas Banerji


IP: 129.82.195.78

We then substitute k = nω/c and v = c/n to obtain: µ ¶ 1 1 dn = n+ω , vg c dω µ ¶ n ω dn = 1+ , c n dω µ ¶ 1 ω dn = 1+ . v n dω The first relationship in eqn 2.26 follows immediately by taking the reciprocal. The second relationship in eqn 2.26 is obtained by substituting λ = 2πc/ω so that dn dn dλ λ2 dn = =− , dω dλ dω 2πc dλ and hence: vg = v[1 + (2πc/λn)(−λ2 /2πc)dn/dλ]−1 = v[1 − (λ/n)dn/dλ]−1 . (2.8) We consider three separate frequency regions. (i) ω < ω0 : In this frequency region ²r is 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. Since ²r > 1, n > 1, and hence v = c/n < c. Therefore vg < c. (ii) ω0 < ω < (ω02 + N e2 /²0 m0 )1/2 : In this frequency region, ²r 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 > (ω02 + N e2 /²0 m0 )1/2 : In this region ²r is positive and increases with frequency, approaching unity asymptotically. dn/dω is therefore positive, 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 2.26. With n = ²r , we obtain: dn d = dω dω

µ ¶1/2 N e2 1 1 N e2 ω 1+ = . 2 2 2 ²0 m0 ω0 − ω n ²0 m0 (ω0 − ω 2 )2

Hence 1 vg

= = =

n N e2 ω2 + , c nc²0 m0 (ω02 − ω 2 )2 µ ¶ N e2 ω2 1 n2 + , nc ²0 m0 (ω02 − ω 2 )2 µ ¶ N e2 1 ω2 ²r + . nc ²0 m0 (ω02 − ω 2 )2

On substituting for ²r from the exercise, we find: µ ¶ 1 1 N e2 ω02 = 1+ , vg nc ²0 m0 (ω02 − ω 2 )2 14


Name: Tapas Banerji


and hence:

IP: 129.82.195.78

µ ¶−1 N e2 ω02 vg = nc 1 + . ²0 m0 (ω02 − ω 2 )2

The denominator is greater than unity, and n < 1, so vg < c. (2.9) (a) Consider a dipole p placed at the origin. The electric field generated at position vector r is given by: E(r) =

3(p · r)r − r2 p . 4π²0 r5

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) − r2 p 3(p · r)r − r2 p = . 4π²0 r5 4π²0 r5

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 r i = (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: pi (3zi2 − ri2 ) Ei = . 4π²0 ri5 We now sum over the cubic lattice of dipoles within the sphere. By symmetry, the x and y components sum to zero, giving a resultant field along the z axis of magnitude: E sphere =

1 X 3zi2 − ri2 , pi 4π²0 i ri5

as required. (b) If all the dipoles have the same magnitude p, then the resultant field is given by: p X 2zi2 − x2i − yi2 E sphere = . 4π²0 i ri5 The x, y and z axes are equivalent for the cubic lattice within the sphere, and so we must have: X y2 X z2 X x2 i i i = = 5 5 5 . r r r i i i i i i It is thus apparent that X 2z 2 − x2 − y 2 i

i

i

ri5

i

= 0.

The net field is therefore zero. (c) Consider a hollow sphere of radius a placed within a polarized dielectric medium as illustrated in Fig. 6. (cf. Fig 2.8.) We assume that the polarization is parallel to the z axis. The surface charge on the sphere must 15


Name: Tapas Banerji


IP: 129.82.195.78

balance the normal component of the polarization P . With P = (0, 0, P ), the normal component at polar angle θ is equal P cos θ, as shown in Fig. 6. Hence the surface charge density σ at angle θ is equal to −P cos θ. The charge contained in a circular element at angle θ subtending an incremental angle dθ as defined in Fig. 6 is then given by: dq = σ dA = −P cos θ × (2πa sin θ · a dθ) = −2πP a2 cos θ sin θ dθ . The x and y components of the field generated at the origin by this incremental charge sum to zero by symmetry, leaving just a z component, with a magnitude given by Coulomb’s law as: dE z = −

dq P cos2 θ sin θ dθ cos θ = + . 4π²0 a2 2²0

On integrating over θ, we then obtain: Z π Z π P P Ez = dE z = cos2 θ sin θ dθ = . 2²0 0 3²0 θ=0 Since P is parallel to the z axis, and the x and y components of E are zero, we therefore have: P E= , 3²0 as required for eqn 2.28.

P

-- - - -

-

+

+

-

a

dq

q

P cosq

dq

+ + +

+

+

Figure 6: Definition of angles and charge increment as required for Exercise 2.9(c). (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 χ a ≡ 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. 16


Name: Tapas Banerji


IP: 129.82.195.78

(2.11) We are working with a gas, and we can therefore forget about ClausiusMossotti. At s.t.p. we have NA (Avogadro’s constant) molecules in a volume of 22.4 litres. Hence N = 2.69 × 1025 m−3 . We then find χa from: χa = (²r − 1)/N = 2.2 × 10−29 m3 . The atomic dipole is worked out from p = ²0 χa E . The displacement of an electron by 1˚ A produces a dipole of 1.6×10−29 C m. 11 Hence we require a field of 0.8×10 V/m. The field acting on an electron at a distance r from a proton is given by Coulomb’s law as: e E= . 4π²0 r2 ˚, we find E = 1.4 × 1011 V/m. It is not surprising On substituting r = 1 A 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) We are given that α(E) = α0 for E2 ≥ E ≥ E1 , where E is the photon energy, and α(E) = 0 at all other energies. By using eqn 1.19 we then have: κ(E) = κ(E) =

c~α0 /2E , E2 ≥ E ≥ E1 , 0 , elsewhere .

We can then find the refractive index from eqn 2.36 as follows: Z 2 E2 /~ ω 0 κ(ω 0 ) n(E) = 1 + dω 0 , π E1 /~ ω 02 − ω 2 Z 2 E2 E 0 κ(E 0 ) = 1+ dE 0 . π E1 E 02 − E 2 Note that we do not have to worry about taking principal parts because κ(E) = 0 when E 0 = E if E < E1 . On substituting for κ(E 0 ) we find: Z c~α0 E2 1 n(E) = 1 + dE 0 . 02 − E 2 π E E1 Now:

Z

1 dx x2 − a2

= =

Hence: n(E)

¸ Z · 1 1 1 − dx , 2a x−a x+a µ ¶ x−a 1 ln . 2a x+a

· µ 0 ¶¸E2 E −E c~α0 1 , ln = 1+ π 2E E 0 + E E1 · µ ¶ µ ¶¸ c~α0 E2 − E E1 − E = 1+ ln − ln , 2πE E2 + E E1 + E µ ¶ c~α0 (E2 − E)(E1 + E) = 1+ ln , 2πE (E2 + E)(E1 − E)

as required. 17


Name: Tapas Banerji


IP: 129.82.195.78

(2.13) (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 e2 X fj 2 − ω2 . ²0 m0 j ω0j

On substituting ω = 2πc/λ, this becomes: n2 = 1 +

N e2 1 X fj λ2j λ2 , ²0 m0 (2πc)2 j λ2 − λ2j

where λj = 2πc/ω0j . This is of the Sellmeier form if we take: Aj = N e2 fj λ2j /4π 2 ²0 m0 c2 . (b) With the approximations stated in the exercise, we have: n2 = 1 +

A1 λ2 = 1 + A1 (1 − λ21 /λ2 )−1 . λ2 − λ21

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 + · · · ] ¶ µ A1 A A21 √ 1 x2 + · · · = (1 + A1 )1/2 + √ x+ − 8(1 + A1 )3/2 2 1 + A1 2 1 + A1

On re-substituting for x, we then find: µ ¶ 4 λ21 A1 A21 A1 λ1 √ + − , n = (1 + A1 )1/2 + √ 2 3/2 λ4 8(1 + A1 ) 2 1 + A1 λ 2 1 + A1 which shows that: C1 C2

= (1 + A1 )1/2 , = A1 λ21 /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 λ21 /λ2 ¿ 1 will be reasonable, as we are far away from the band gap energy.

18


Name: Tapas Banerji


IP: 129.82.195.78

(2.14) (a) On neglecting the λ4 term in Cauchy’s formula, we have n = C1 + C2 /λ2 . On inserting the values of n at 402.6 nm and 706.5 nm and solving, we 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. (b) The values are found by substituting into the result found in part (a) to obtain n = 1.5493 at 450 nm and n = 1.5369 at 650 nm. (c) Referring to the angles defined in Fig. 7, we have sin θin sin θout = = n, sin θ1 sin θ2 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◦ .

60° qin qout q1

q2

Figure 7: Angles required for the solution of Exercise 2.14. (2.15) The transit time is given by: τ=

L dk =L , vg dω

which, with k = nω/c, becomes: τ=

L c

µ ¶ dn n+ω . dω

We introduce the vacuum wavelength λ via ω = 2πc/λ, and write: dn dn dλ λ2 dn = · =− . dω dλ dω 2πc dλ We then have (cf. eqn 2.26 with τ = L/vg ): µ ¶ L dn τ= n−λ . c dλ The difference in the transit time for two wavelengths separated by ∆λ, where ∆λ ¿ λ, is given by: ∆τ =

dτ ∆λ . dλ

19


Name: Tapas Banerji


IP: 129.82.195.78

On using the result for τ above, we find: L d2 n dτ =− λ 2, dλ c dλ which implies: ∆τ = L

µ ¶ λ d2 n − ∆λ = L D ∆λ , c dλ2

where D is the material dispersion parameter defined in eqn 2.40. We therefore finally obtain: ¯ ¯ ¯ λ d2 n ¯ ¯ ¯ ∆λ = L |D| ∆λ . |∆τ | = L ¯− c dλ2 ¯ The time–bandwidth product of eqn 2.39 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. The precise amount of broadening depends on the numerical value of the time–bandwidth product assumed for the pulse. For a 10 ps pulse with ∆ν∆t = 1, we have ∆ν = 1011 Hz. With λ = c/ν, we have |∆λ| = (λ2 /c)|∆ν|, so that |∆λ| = 0.8 nm in this case. We thus find: |∆τ | = L|D|∆λ = 1 km × 17 ps km−1 nm−1 × 0.8nm = 14 ps .

e-ray polarization vector

z

direction of propagation

ne q

n(q ) no

no

y

index ellipsoid ne Figure 8: 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 2.16. (2.16) It is apparent from eqn 2.50 that ²11 /²0 = ²22 /²0 = n2o and ²33 /²0 = n2e . Hence we can re-cast the index ellipsoid in the form: x2 y2 z2 + 2 + 2 = 1. 2 no no ne Owing to the x-y symmetry about the optic (z) axis, we can choose the axes of the index ellipsoid so that the x axis coincides with the polarization 20


Name: Tapas Banerji


IP: 129.82.195.78

vector of the o–ray as in Fig 2.13(a). The polarization of the e–ray will then lie in the y–z plane, as in Fig. 2.13(b). The projection of the index ellipsoid onto the plane that contains the direction of propagation and the polarization vector of the e–ray thus appears as the ellipse drawn in Fig. 8. The refractive index n(θ) that we require is the distance from the 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)2 cos2 θ n(θ)2 sin2 θ + = 1, n2o n2e

which implies:

1 cos2 θ sin2 θ = + , n(θ)2 n2o n2e

as required. (2.17) The condition for total internal reflection for a medium–air interface is that the angle of incidence should exceed the critical angle θc given by: sin θc = 1/n , where n is the refractive index of the medium. We see from Table 2.1 that no = 1.658 and ne = 1.486 for calcite. The critical angles for the o- and e-rays are therefore 37.1◦ and 42.3◦ respectively. For the polarizer to work, we want the o-ray to suffer total internal reflection, but not the e-ray. With normal incidence as shown in Fig. 2.14, this will occur if the apex angle θ lies in the range 37.1◦ − 42.3◦ .

y

y Input E-vector

(a)

Input E-vector direction

(b)

Ey

q Ez

Ez

optic axis (z)

q -Ey

front surface of wave plate

optic axis (z)

Output E-vector

Figure 9: (a) Input polarization vector for light propagating along the x direction, relative to the crystal axes, as required for the solution of Exercise 2.18(a). (b) Output polarization of the half-wave plate. (2.18) (a) We resolve the input polarization into two components, one along the optic axis, and one orthogonal to it, as shown in Fig. 2.15(b). We define our axes so that z lies along the optic axis, and y lies in the plane of the front surface of the waveplate, as shown in Fig. 9(a). The x axis is taken

21


Name: Tapas Banerji


IP: 129.82.195.78

to be the direction of propagation. If the input beam has amplitude E 0 , the input polarization has components: E in z E in y

= E 0 cos θ , = E 0 sin θ .

The component along z is the e-ray and the one along y is the o-ray. In a half-wave plate, the phase of the o-ray is shifted by π relative to the e-ray at the output of the retarder plate. This means that the output polarizations will be given by: E out z E out y

= E 0 cos θ = E in z , = E 0 eiπ sin θ = −E 0 sin θ = −E in y ,

as shown in Fig. 9(b). The resultant E-vector points at an angle −θ with respect to the optic axis. Hence the output polarization is rotated by an angle 2θ with respect to the input polarization. (b) If we set θ = 45◦ , the input polarization will be given by: √ E in = E 0 cos 45◦ = E 0 / 2 , z √ E in = E 0 sin 45◦ = E 0 / 2 . y The quarter-wave plate introduces a phase difference of π/2 between the e-ray and o-rays. The output polarization will therefore be: √ E out = E 0/ 2 , z √ √ E out = (E 0 / 2)eiπ/2 = i E 0 / 2 . y The output therefore consists of two orthogonal linearly polarized components of equal amplitude with a phase difference of 90◦ between them. This is circularly-polarized light. (c) When θ 6= 45◦ , the amplitudes of the e- and o-rays will be different. The quarter-wave plate still introduces a phase shift of 90◦ between them, and so the output consists of two orthogonal linearly polarized components with unequal amplitude and with a 90◦ phase difference between them. This is elliptically polarized light. (2.19) We follow Example 2.4, but set the phase difference ∆φ to be π/2 because we have a quarter-wave plate instead of a half-wave plate. We therefore require: 2π|∆n|d π , ∆φ = = 2 λ which implies d = λ/4|∆n|. In this case we have |∆n| = 0.0091 and λ = 500 nm, and so we require d = 1.4 × 10−5 m. In this exercise we have neglected the optical activity of the quartz plate because the thickness is very small. In a thicker multiple-order waveplate (i.e one with ∆φ = (2πm + π/2), where m is an integer) a small correction would have to be introduced to account for the optical activity.

22


Name: Tapas Banerji


IP: 129.82.195.78

(2.20) 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 (a) NaCl (b) Diamond (c) Graphite (d) Wurtzite (e) Zinc blende (f) Solid argon (g) Sulphur

Structure cubic (fcc) cubic hexagonal hexagonal cubic cubic (fcc) orthorhombic

x, y, z equivalent ? yes yes no no yes yes no

Birefringent ? no no yes yes no 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. (2.21) (a) The birefringent phase shift for a refractive difference of ∆n is given by eqn 2.46. For a half wavelength shift, this is equal to π. On substituting for the Kerr-induced birefringence from eqn 2.51, we obtain: φ=π=

2π|∆n|d 2π(λKE 2 )d = = 2πKE 2 d . λ λ

On solving for the field, we then find: √ E λ/2 = 1/ 2Kd , as required. (b) We substitute for K and d to find: p E λ/2 = 1/ 2 × (8.7 × 10−14 ) × 0.02 = 1.7 × 107 V/m . The field is dropped across a distance of 5 mm, and so the voltage required is Vλ/2 = E λ/2 × 0.005 = 85 kV. As Table 2.2 shows, the value of K for the chalcogenide considered here is large for a glass, but the voltage required to make a Kerr cell is still very high. Practical Kerr cells use liquids with larger Kerr constants, for example, nitrobenzene. (2.22) (a) Set up axes so that z corresponds to the direction of propagation and x to the input polarization. With these definitions the input polarization can be written as: E in = E 0 x ˆ, where E 0 is the amplitude of the light. Circular polarization consists of two orthogonally polarized waves of equal amplitude with a 90◦ phase difference between them, and so we can write left and right circular light as: √ σ + = E 0 (ˆ x + iˆ y)/ 2 , √ σ − = E 0 (ˆ x − iˆ y)/ 2 , 23


Name: Tapas Banerji


IP: 129.82.195.78

where the factor of i ≡ eiπ/2 represents the 90◦ phase shift. This allows us to re-write the input polarization in terms of the left and right circular light as: √ E in = E 0 (σ + + σ − )/ 2 . An optically-active medium will introduce a phase difference φ between the left- and right-circular components, where: φ=

2π ∆n d . λ

Here, λ is the vacuum wavelength, ∆n = (nR − nL ) is the difference of the refractive indices for right- and left-circular light, and d is the thickness of the medium. The output wave will therefore be of the form: √ E out = E 0 (σ + + eiφ σ − )/ 2 . We revert to the linear basis by substituting for σ + and σ − and re-writing this as: ¡ ¢ E out = E 0 (ˆ x + iˆ y) + eiφ (ˆ x − iˆ y) /2 . Collecting the terms in x ˆ and y ˆ, we find: ¡ ¢ E out = E 0 (1 + eiφ )ˆ x + i(1 − eiφ )ˆ y) /2 , ³ ´ = E 0 eiφ/2 (e−iφ/2 + eiφ/2 )ˆ x + i(e−iφ/2 − eiφ/2 )ˆ y) /2 , = =

E 0 eiφ/2 (2 cos(φ/2) x ˆ + i(−2i sin(φ/2)) y ˆ) /2 , E 0 eiφ/2 (cos(φ/2) x ˆ + sin(φ/2) y ˆ) .

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

φ π πd = ∆n d = (nR − nL ) . 2 λ λ

(b) Using the result from part(a), we find that the rotatory power of an optically-active medium is given by: RP = |θ/d| = π|∆n|/λ . For quartz we thus find RP = π(7.1 × 10−5 )/589 × 10−9 = 379 rad/m = 21.7 ◦ /mm . (2.23) We apply a Lorentzian model according to the theory in Section 2.2. The Lorentzian lineshapes are shifted for σ + and σ − light due to the Zeeman effect, which shifts the energy of the transition from ~ω0 to ~ω0 ± µB B according to the circular polarization. (a) The Faraday rotation is caused by the difference of the real part of the refractive index for σ + and σ − light. Figure 10(a) plots the refractive indices for two Lorentzian lines shifted from each other by 2µB B and their 24



IP: 129.82.195.78

1.04 0.04

Dn

1.02

0.02

1.00

0.00

0.98

nn+

-0.02

Dn

-0.04

Dn

Refractive index n

(a)

0.96 0.6

0.8

1.0

1.2

1.4

Angular frequency (w/w0) 0.06

Extinction coefficient k

Name: Tapas Banerji

0.04

kk+

(b)

0.02

Dk 0.00 -0.02 -0.04

Dk

-0.06 0.6

0.8

1.0

1.2

1.4

Angular frequency (w/w0) Figure 10: Lorentzian model calculation of Faraday effect and magnetic circular dichroism, as considered in Exercise 2.23. (a) Faraday rotation. (b) Magnetic circular dichroism. difference. The lines are plotted against the normalized angular frequency (ω/ω0 ) for a Lorentzian line with γ = 0.05ω0 . The value of µB B/~ is set to be equal to γ. The value of ∆n = Re(˜ n+ − n ˜ − ), and hence the Faraday rotation is negative above and below the line, and positive near ω0 . The rotation decays as the frequency is tuned away from resonance. (b) The magnetic circular polarization is found by calculating the difference ∆κ of the imaginary part of (˜ n+ − n ˜ − ). This is shown in Fig. 10 for the same Lorentzian line as in part (a). We see that the magnetic circular dichroism follows a dispersive lineshape, with a negative signal below ω0 that peaks at ω0 − µB B/~, and a positive signal above ω0 that peaks at ω0 + µB B/~. The signal precisely at ω0 is zero. (2.24) We require a Faraday rotation of π/4 at a magnetic field of 0.5 T. The Faraday rotation is given by eqn 2.53, and so the thickness of glass is given

25


Name: Tapas Banerji


IP: 129.82.195.78

by:

θ π/4 = = 0.17 m = 17 cm . Vb 9.0 × 0.5 This long length is impractical, and real Faraday isolators use specialist glasses or crystals with larger Verdet coefficients. d=

26


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 3 Interband absorption (3.1) The Born–von Karman boundary conditions are satisfied when: kx L ky L kz L

= = =

nx L , ny L , nz L ,

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. 11. 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)3 states per unit volume of k space. In most calculations we actually require the density of states in k space as given in eqn 3.15. This 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. 11. This volume of the incremental shell is equal to 4πk 2 dk, and contains 4πk 2 dk × 1/(2π)3 = k 2 dk/2π 2 states. Hence g(k) = k 2 /2π 2 . ky dk

k kx

2p/L

Figure 11: Grid of allowed values of k permitted by the Born–von Karman boundary conditions, as considered in Exercise 3.1. The points of the grid are separated 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.

27


Name: Tapas Banerji


IP: 129.82.195.78

(3.2) With E(k) = ~2 k 2 /2m∗ , we have: ~2 k dE = ∗ . dk m On inserting into eqn 3.14–15 and substituting for k, we find: g(E) = 2

k 2 /2π 2 m∗ k m∗ = 2 2 = 2 2 2 ∗ ~ k/m π ~ π ~

µ

2m∗ E ~2

¶1/2 =

1 2π 2

µ

2m∗ ~2

¶3/2 E 1/2 ,

as required. (3.3) (a) 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 symmetry 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 definite parity. (b) r is an odd function, and so the integral over all space will be zero unless the product ψf∗ ψi is 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. (c) In spherical polar co-ordinates (r, θ, φ) we have: x = y = z =

r sin θ cos φ = r sin θ (eiφ + e−iφ )/2 , r sin θ sin φ = r sin θ (eiφ − e−iφ )/2i , 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: Z 2π 0 M∝ e−im φ · 1 · eimφ dφ , φ=0

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: Z 2π 0 M∝ e−im φ (eiφ ± e−iφ ) eimφ dφ , φ=0

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

√ = E 0 (ˆ x + iˆ y)/ 2 , √ = E 0 (ˆ x − iˆ y)/ 2 ,

28


Name: Tapas Banerji


IP: 129.82.195.78

where E ± are the electric fields for σ ± polarizations. With x = r sin θ cos φ and y = r sin θ sin φ, we therefore find: √ E + = E 0 r sin θ(cos φ + i sin φ)/ 2 ∝ e+iφ , √ E − = E 0 r sin θ(cos φ − i sin φ)/ 2 ∝ e−iφ . On inserting these into the matrix element, we have for σ + light: Z 2π 0 M+ ∝ e−im φ e+iφ eimφ dφ , φ=0

which is zero unless m0 = m + 1. Similarly, for σ − light we have: Z 2π 0 M− ∝ e−im φ e−iφ eimφ dφ , φ=0

which is zero unless m0 = m − 1. With circularly polarized light we therefore have ∆m = +1 for σ + polarization and ∆m = −1 for σ − polarization. (3.4) The apparatus required is basically the same as for Figs 3.14–15, but 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 means that a detector with a band gap smaller than 0.3 eV must be used, e.g. InSb. (See Table 3.3.) As InSb array detectors are not available, a 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 12 gives a diagram of a typical arrangement that could be used.

computer white light source

InAs sample

: InSb detector

infrared lenses

scanning monochromator

Figure 12: 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 29



IP: 129.82.195.78

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 absorption mechanisms will freeze out as the temperature is lowered. (3.6) Plot α2 and α1/2 against ~ω as shown in Fig. 13. In the range 2.2 ≤ ~ω ≤ 2.7 eV, the graph of α2 is a straight line with an intercept at 2.2 eV. We thus 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, which implies that GaP has a direct gap at 2.75 eV. The huge difference in the two axis scales of Fig. 13 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 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. 1000

80 60

600 40

400

20

200 0

a2 (1012 m-2)

800 a1/2 (m-1/2)

Name: Tapas Banerji

2.2

2.4

2.6 2.8 Energy (eV)

3.0

0

Figure 13: Analysis of GaP absorption data as required for Exercise 3.6. (3.7) A wavelength of 1200 nm corresponds to a photon energy of 1.03 eV. This is above the direct gap of germanium at 0.805 eV, and thus the absorption will be given by (cf. eqn 3.25): α = C(~ω − 0.805)1/2 . The scaling coefficient C can be determined from the data in Fig. 3.11: α ≈ 6 × 105 m−1 at 0.86 eV implies that C ≈ 2.5 × 106 m−1 eV−1/2 . With this value of C, we then find α ≈ 1.2 × 106 m−1 at 1.03 eV. (3.8) (a) We consider transitions 1 and 2 in Fig. 3.5. The k vectors for the transitions can be worked out from eqn 3.23. The appropriate parameters are read from Table D.2 as follows: Eg = 1.424 eV, m∗e = 0.067me , m∗hh = 0.5me , and m∗lh = 0.08me . For the heavy-hole and light-hole transitions we find from eqn 3.22 that µhh = 0.059 me and µlh = 0.036 me respectively. Hence for ~ω = 1.60 eV, we find k = 5.3 × 108 m−1 for the heavy holes and k = 4.1 × 108 m−1 for the light holes. (b) 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 30


Name: Tapas Banerji


IP: 129.82.195.78

crystal is therefore given by: k=

2π = 3.0 × 107 m−1 . (λ/n)

This is more than an order of magnitude smaller than the electron wave vector, and hence the approximation in eqn 3.12 is justified. (c) 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 . (d) 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) Consider a σ + transition. By referring to Fig. 3.8, we see that in a heavyhole transition we start from the |J, MJ i = |3/2, −3/2i sublevel. For heavy holes we must therefore insert J = 3/2 and MJ = −3/2 into the matrix element for σ + transitions. This gives: |hJ − 1, MJ + 1|σ + |J, MJ i|2

1 2 1 2

= =

(J − MJ )(J − MJ − 1)C ,

(3/2 − (−3/2))(3/2 − (−3/2) − 1)C , = 3C. For light holes we start from the |J, MJ i = |3/2, −1/2i sublevel in a σ + transition, and so the matrix element is given by: |hJ − 1, MJ + 1|σ + |J, MJ i|2

= 21 (J − MJ )(J − MJ − 1)C , = 21 (3/2 − (−1/2))(3/2 − (−1/2) − 1)C , = C.

The ratio of the squares of the matrix elements is therefore 3C : C = 3 : 1. The argument is the same for σ − transitions. For heavy holes we start from the |J, MJ i = |3/2, +3/2i sublevel, and so the heavy-hole matrix element is given by: |hJ − 1, MJ − 1|σ − |J, MJ i|2

1 2 1 2

=

(J + MJ )(J + MJ − 1)C , (3/2 + (+3/2))(3/2 + (+3/2) − 1)C , = = 3C.

For light holes we have |J, MJ i = |3/2, +1/2i, and hence: |hJ − 1, MJ − 1|σ − |J, MJ i|2

1 2 1 2

= = =

(J + MJ )(J + MJ − 1)C , (3/2 + (+1/2))(3/2 + (+1/2) − 1)C ,

C.

The ratio is therefore 3 : 1 again.

31


Name: Tapas Banerji


IP: 129.82.195.78

(3.10) Linearly polarized light can be considered as a superposition of two oppositely circularly polarized waves. (See eqn A.41.) The σ + part of the wave will create negative spin polarization, while the σ − part will create an identical positive spin polarization. Hence the net spin polarization is always equal to zero. (3.11) The main issue to think about is the effect of the split-off hole band. A split-off hole transition with σ + polarization will generate a spin up electron, and thus start to cancel the negative spin polarization created by the heavy-hole transitions. The matrix element squared for σ ± split-off holes transitions is twice that of the light-hole transitions, [see eg Meier & Zacharchenya (1984), p.24] so that the combined strength of the (light-hole + split-off hole) transitions is equal to that of the heavy-hole transition. This means that the net spin polarization will ultimately go to zero when split-off hole transitions become significant. Thus we expect that the electron spin polarization will be constant at ±50% for Eg ≤ ~ω ≤ (Eg + ∆), where ∆ is the spin–orbit energy, and then to drop for higher photon energies. This is in fact observed experimentally. The spin polarization is effectively zero for ~ω À (Eg + ∆). One way to understand this is to realize that the electric field of the light only interacts with the orbital motion of the electron and does not interact directly with its spin. Therefore, if spin polarization is being generated optically, it must be through the spin–orbit interaction. Hence we expect the effect to disappear when the electron energy is larger than the spin–orbit energy ∆. (3.12) (a) The lowest conduction band states of silicon are p–like at the Γ point (i.e k = 0). The spin–orbit splitting is small, and the J = 1/2 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. (b) 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. (3.13) At low temperatures, phonon absorption is impossible, and the indirect transition must proceed by phonon emission, with a threshold energy of Egind + ~Ω. The band structure diagram of germanium given in Fig. 3.10 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 Egind + 0.008 eV, i.e. at 0.75 eV, as indeed demonstrated by the data in Fig. 3.11(a). (3.14) The absorption coefficient with a field applied is given by eqn 3.26. The absorption decreases exponentially for ~ω < Eg , and this produces an exponential absorption tail below Eg . Although there is no clear absorption 32


Name: Tapas Banerji


IP: 129.82.195.78

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 Eg . This occurs when √ 4 2m∗e (Eg − ~ω)3/2 = 1 . 3|e|~E 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 (0.01 eV)3/2 = 1 . 3|e|~E With m∗e = 0.067me , we find E = 1.8 × 106 V/m.

v F B

w r

Figure 14: Force acting on a particle moving in a magnetic field pointing into the paper, as required for Exercise 3.15. The charge is assumed to be positive. (3.15) We consider a particle of charge q, mass m and velocity v moving in 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. 14. 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: mω 2 r = 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) eikz z , 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

33



IP: 129.82.195.78

modified wave functions. For x-polarized light, the matrix element is now given by: Z 0 M ∝ u∗f (r)ϕn0 (x, y)e−ikz z x u∗i (r)ϕn (x, y)e−ikz z d3 r , Z u∗f (r) x u∗i (r) d3 r × hϕn0 |ϕn i , ∝ unit cell

where we assumed kz0 = kz as usual in the second line. The electric dipole matrix element is therefore proportional to the overlap of harmonic oscillator wave functions with different values of n. Now harmonic oscillator 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. (3.16) (a) Let z be the free direction. Apply Born–von Karman boundary conditions 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 = ~2 k 2 /2m for a particle of mass m, which implies: p dE/dk = ~2 k/m = ~ 2E/m . The density of states in energy space is then worked out from eqn 3.14: g1D (E) = 2

1/2π g(k) =2 p = (2m/Eh2 )−1/2 . dE/dk ~ 2E/m

We thus see that g1D (E) ∝ E −1/2 . (b) The threshold energy will be equal to the band gap Eg of the semiconductor as for a 3-D material. Fermi’s golden rule indicates that the 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. 15(i). 6

Absorption (a.u.)

Absorption (a.u.)

Name: Tapas Banerji

(i) 4 2 0

6 (ii) 4 2 0

0 2 4 6 8 10 Energy relative to Eg in arb. units

0 1 2 3 Energy in units of h wL relative to Eg

Figure 15: (i) Absorption of a one-dimensional semiconductor as discussed in Exercise 3.16(b). (ii) Absorption for a system with quantized Landau levels in two directions and free motion in the third. ωL is the Landau level angular frequency, as discussed in Exercise 3.16(c). (c) The magnetic field quantizes the motion in two dimensions to give Landau levels, leaving the particle free to move in the third dimension. 34


Name: Tapas Banerji


IP: 129.82.195.78

Optical transitions are possible between Landau levels with the same value of n. (See Exercise 3.15.) These will occur at energies given by (cf. 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. 15(ii). 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. (d) 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∗h À 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. 3.10.) The indirect transitions for ~ω > 0.66 eV are too weak to be observed compared to the direct transitions above 0.80 eV. (3.17) The responsivity is calculated by using eqn 3.39. The device will be most efficient if it has 100% quantum efficiency, and so we set η = 1, giving: Responsivitymax =

e (1 − e−αl ) . ~ω

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. (3.18) (a) The p–i–n diode structure is described in Appendix E. 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. (b) The capacitance of a parallel-plate capacitor of area A, permittivity ²r ²0 and plate separation d is given by: C=

A²r ²0 . 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 = 10−6 m2 , ²r = 11.9 and li = 10−5 m. 35


Name: Tapas Banerji


IP: 129.82.195.78

(c) The electric field for an applied bias of −10 V can be calculated from eqn E.3 as: 1.1 − (−10) = 1.1 × 106 V/m . E= 10−5 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. (d) 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 × 104 m/s . This velocity occurs for an electric field of v/µe = 1.3 × 105 V/m. We finally find the voltage from eqn E.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.

36


Name: Tapas Banerji


IP: 129.82.195.78

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 r 1 and r 2 , the Hamiltonian thus takes the form: 2 2 e2 ˆ = − ~ ∇2 − ~ ∇2 − H , 2m1 1 2m2 2 4π²0 |r 1 − r 2 |

where m1 = me , m2 = mp , and ∇2i =

∂2 ∂2 ∂2 + + . ∂x2i ∂yi2 ∂zi2

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

= r1 − r2 m1 r 1 + m2 r 2 = . m1 + m2

Now ∂ ∂x1 ∂ ∂x2

= =

∂x ∂ ∂X ∂ ∂ m1 ∂ + = + ∂x1 ∂x ∂x1 ∂X ∂x m1 + m2 ∂X ∂x ∂ ∂X ∂ ∂ m2 ∂ + =− + , ∂x2 ∂x ∂x2 ∂X ∂x m1 + m2 ∂X

which implies ∂2ψ ∂x21

=

∂2ψ 2m1 ∂2ψ + + ∂x2 m1 + m2 ∂x∂X

∂2ψ ∂x21

=

∂2ψ 2m2 ∂2ψ − + 2 ∂x m1 + m2 ∂x∂X

µ µ

m1 m1 + m2 m2 m1 + m2

¶2 ¶2

∂2ψ ∂X 2 ∂2ψ . ∂X 2

It is then apparent that: 1 ∂2ψ 1 ∂2ψ + = 2 m1 ∂x1 m2 ∂x22

µ

1 1 + m1 m2

¶

∂2ψ 1 ∂2ψ + . 2 ∂x m1 + m2 ∂X 2

Similar results may be derived for the other co-ordinates, so that we have: µ ¶ 1 2 1 1 1 1 2 ∇1 + ∇2 = + ∇2r + ∇2 m1 m2 m1 m2 m1 + m2 R

37


Name: Tapas Banerji


IP: 129.82.195.78

On introducing the total mass M and reduced mass µ according to: M 1 µ

= =

m1 + m2 , 1 1 + , m1 m2

and substituting into the Hamiltonian, we then find 2 2 2 ˆ = − ~ ∇2R − ~ ∇2r − e . H 2M 2µ 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 ˆ whole atom + H ˆ relative , H where: ˆ whole atom H ˆ relative H

~2 2 ∇ 2M R ~2 e2 = − ∇2r − . 2µ 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 µ 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 µ, rather than the individual electron mass. The reduced mass correction does not make much difference for hydrogen itself, where m2 À m1 and hence µ ≈ m1 , but it is very important for excitons, where the electron and hole masses are typically of the same order of magnitude. (4.2) (a) The Hamiltonian describes the relative motion of the electron and hole as they experience their mutual Coulomb attraction within the semiconductor. The kinetic energy of the exciton as a whole is not included. As explained in Exercise 4.1, the appropriate mass is the reduced electronhole 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 ²r in the Coulomb terms accounts for the relative permittivity of the semiconductor. (b) We substitute Ψ into the Schrödinger equation with the ∇2 operator written in spherical polar co-ordinates: µ ¶ µ ¶ 1 ∂ ∂ 1 ∂2 1 ∂ 2 ∂ 2 r + 2 sin θ + 2 2 . ∇ = 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 38


Name: Tapas Banerji


IP: 129.82.195.78

Since Ψ does not depend on θ or φ, the Schrödinger equation becomes: µ ¶ ~2 1 ∂ e2 2 ∂Ψ − r − Ψ = E Ψ. 2 2µ r ∂r ∂r 4π²0 ²r r On substituting Ψ = C exp(−r/a0 ), we obtain: ¶ µ ~2 e2 ~2 + − Ψ = E Ψ. − 2µa20 µa0 r 4π²0 ²r r The wave function is therefore a solution if +

e2 ~2 − = 0, µa0 r 4π²0 ²r r

which implies

4π²0 ²r ~2 m0 ≡ ²r aH , 2 µe µ is the hydrogen Bohr radius. With this value of a0 , we then a0 =

where aH find:

µe4 ~2 µ 1 =− 2 2 2 ≡− RH , 2 2µa0 8²r ²0 h m0 ²2r where RH is the hydrogen Rydberg energy. E=−

The normalization constant is found by solving: Z ∞ Z π Z 2π Ψ∗ Ψ r2 sin θ drdθdφ = 1 . r=0

This gives:

θ=0

Z 4πC 2

∞

φ=0

r2 e−2r/a0 dr = 4πC 2 ×

r=0

which implies:

µ C=

1 πa30

a30 = 1, 4

¶1/2 .

In the language of atomic physics, the wave function considered here is 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) ∝ r2 e−2r/a0 . On differentiating, we find that this peaks at a0 . The expectation value of r is found from Z ∞ Z π Z 2π hri = Ψ∗ rΨ r2 sin θ drdθdφ , r=0 θ=0 φ=0 µ ¶Z ∞ 1 = 4π r3 e−2r/a0 dr , πa30 r=0 = (3/2)a0 . The peak of the probability density and the expectation value thus differ by a factor of 3/2. 39


Name: Tapas Banerji


IP: 129.82.195.78

(4.4) The purpose of this exercise is to familiarize the student with the variational 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.) (a) 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. (b) We substitute Ψ into the Hamiltonian with the ∇2 written in spherical polar co-ordinates, as in Exercise 4.2. Since Ψ again depends only on r, this gives: µ ¶ ~2 1 ∂ e2 2 ∂Ψ ˆ HΨ = − r − Ψ. 2µ r2 ∂r ∂r 4π²0 ²r r On evaluating the derivatives, we find: · µ 2 ¶ ¸µ ¶1/2 2 2 1 1 ˆ = − ~ + ~ − e HΨ e−r/ξ . 2µξ 2 µξ 4π²0 ²r r πξ 3 The expectation value is then given by: Z ∞ Z π Z 2π ˆ r2 sin θ drdθdφ hEi = Ψ∗ HΨ r=0

µ

θ=0

1 πξ 3

φ=0 ∞

¶Z

·

=

4π

−

=

~2 e2 − . 2 2µξ 4π²r ²0 ξ

r=0

~2 2 r + 2µξ 2

µ

~2 e2 − µξ 4π²0 ²r

¶ ¸ r e−2r/ξ dr

(c) On differentiating hEi with respect to ξ, we find a minimum when ξ = 4π²0 ²r ~2 /µe2 . The value of hEi at this minimum is hEi = −µe4 /8h2 ²20 ²2r . (d) 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 approximation 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) (a) The electron performs circular motion around the nucleus with quantized angular momentum equal to n~. The orbits are stable, and photons are only emitted or absorbed when the electron jumps between orbits. (b) The central force for the circular motion is provided by the Coulomb attraction, and the electron velocity v must therefore satisfy: µv 2 e2 = , r 4π²0 ²r r2

40


Name: Tapas Banerji


IP: 129.82.195.78

where r is the radius of the orbit and µ is the reduced mass. (See Exercises 4.1 and 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=

4π²0 ²r ~2 2 m0 ²r n2 aH , n ≡ 2 µe µ

where aH = 4π²0 ~2 /m0 e2 is the hydrogen Bohr radius. The energy is found from: 1 e2 . E = µv 2 − 2 4π²0 ²r r On solving for v and substituting, we find: E=−

µe4 µ 1 RH ≡− , 8h2 ²20 ²2r n2 m0 ²2r n2

where RH = m0 e4 /8h2 ²20 is the hydrogen Rydberg energy. These are the same results as in eqns 4.1 and 4.2. (c) E is identical to the exact solution of the hydrogen Schrödinger equation. (d) 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) > kB T . At room temperature kB T = 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 2.15– 21. At line centre, we have from eqns 1.19 and 1.28: α=

4π²2 4πκ = , λ 2nλ 41


Name: Tapas Banerji


IP: 129.82.195.78

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 1.5149 ω0 = (²st − ²∞ ) , ²2 (ω0 ) = 1.37 = (²st − ²∞ ) γ 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 1.25 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 conclude 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 separated by a distance of r in a medium of relative permittivity ²r is given by Coulomb’s law as: e E= . 4π²0 ²r r2 In the Bohr model we have (see e.g. Exercise 4.5): |E(n)| =

µe4 8(²0 ²r hn)2

and

4π²0 ²r ~2 n2 . µe2 It is thus apparent that for n = 1 we have: µ ¶2 2|E(1)| 2RX e πµe2 e = = = . er1 eaX 4π²0 ²r ²0 ²r h2 4π²0 ²r r2 rn =

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 × 104 V/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 × 10−6 m, we find E = 9.7 × 104 V/m for V0 = +0.55 V. Hence we need to apply a forward voltage of 0.55 V. The excitons will therefore only be observed for forward bias voltages above about 0.5 V. At zero bias and in reverse bias, the excitons will be ionized. (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: µRH e~B = , µ m0 ²2r 42


Name: Tapas Banerji


IP: 129.82.195.78

which, on solving for B, gives: B=

µ2 RH . m0 ²2r e~

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 = (B/2)(−y, x, 0), we obtain:       ∂ 0 −y B x   ∂y x = 0 . B= × 2 ∂z 0 B 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 = (B/2)(−y, x, 0) is given by: 2 2 2 2 ˆ 0 = e A = e B (x2 + y 2 ) . H 2m0 8m0

The diamagnetic energy shift is calculated from 2 2 ˆ 0 |ψi = e B hψ|(x2 + y 2 )|ψi . δE = hψ|H 8m0

In the case of an exciton, the wave functions are spherically symmetric so that: hx2 i = hy 2 i = hz 2 i = hr2 i/3 = rn2 /3 . The total shift is obtained by summing the energy shifts of the electrons and holes to obtain: δE =

e2 B 2 2rn2 e2 B 2 2rn2 e2 B 2 rn2 + = . ∗ 8m∗e 3 8mh 3 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 × 10−5 eV at B = 1 T. The wavelength shift can be calculated from: δλ =

hc dλ δE = − 2 δE = −0.026 nm . dE E

(4.15) The effective masses given imply that 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 for ²r = 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 excitons respectively. 43


Name: Tapas Banerji


IP: 129.82.195.78

(4.16) The classical theory of equipartition of energy states that we have a thermal energy of kB T /2 per degree of freedom. A free particle has three degrees of freedom corresponding to the three velocity components. Hence the total thermal kinetic energy is given by: 3 p2 = kB T . 2m 2 This implies a thermal de Broglie wavelength of: λdeB =

h h =√ . p 3mkB T

The particle density at the Bose–Einstein condensation temperature is given by eqn 4.9. On setting N = 1/r3 , where r is the average interparticle distance, we find √ 1 1/3 2πmkB T = (2.612) . r h Hence: r λdeB

=

1 (2.612)1/3

r

3 = 0.50 . 2π

(4.17) Bose–Einstein condensation refers to the quantization of the kinetic energy due to free translational motion. In applying this to excitons, we need to consider the centre of mass motion of the whole exciton, which behaves as a composite boson. The appropriate mass to use to calculate the condensation temperature is therefore the total mass of the exciton, namely (m∗e + m∗h ) = 1.7m0 . On substituting into eqn 4.9 with N = 1 × 1024 m−3 and solving for Tc , we find Tc = 17.2 K. (4.18) The excitonic radii calculated from eqn 4.2 are r1 = 0.85 nm and r2 = 3.4 nm. For the n = 1 exciton, the radius is comparable to the unit cell size a and thus the Wannier model is invalid. On the other hand, the n = 2 exciton satisfies the condition r À a, and the Wannier model is valid.

44


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 5 Luminescence (5.1) Indirect transitions involve the absorption or emission of a phonon to conserve 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 bottom of the conduction band by emission of phonons, and similarly for holes. The relaxation processes occur on a much faster timescale (∼ps) than the radiative recombination (∼ns). Hence all the carriers have relaxed before emission occurs, and it makes no difference where they were initially injected. 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 Einstein A coefficient is therefore given from eqn B.31 as: e2 ω 3 1 X A= |h2p, m|r|1si|2 , 3π²0 ~c3 3 m=−1,0,1 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 they cancel out.) We write: ˆ, r = xî + yˆj + z k so that:

|hri|2 = |hxi|2 + |hyi|2 + |hzi|2 .

Now atoms are spherically symmetric, and so it must be the case that: |hxi|2 = |hyi|2 = |hzi|2 . We therefore only have to evaluate one of these, and we chose hzi because the mathematics is easier. Written explicitly, with z = r cos θ, we have: Z ∞ Z π Z 2π hzi = Ψ∗2p r cos θ Ψ1s r2 sin θ drdθdφ . r=0

θ=0

φ=0

This has to be evaluated for each of the three possible m values of the 2p state. However, since z has no dependence on φ, and the φ dependence of the wave functions is determined only by m, the integral is only non-zero for the m = 0 level of the 2p state. Hence we only have to evaluate one integral. On inserting the explicit forms of the wave functions, we then have: Z ∞ Z π Z 2π ∗ 2 ∗ hzi = R21 rR10 r dr Y1,0 cos θ Y0,0 sin θ dθdφ , r=0

=

√

1 6a4H

Z

θ=0

∞

φ=0

r4 exp(−3r/2aH ) dr ×

r=0

√ Z π Z 2π 3 cos2 θ sin θ dθdφ . 4π θ=0 φ=0

45


Name: Tapas Banerji


IP: 129.82.195.78

This gives: 24 hzi = √ 6

µ ¶5 2 1 aH × √ = 0.745 aH = 3.94 × 10−11 m . 3 3

Then, with |hri|2 = 3|hzi|2 , we obtain: A=

e2 ω 3 1 e2 ω 3 2 3|hzi| = |hzi|2 . 3π²0 ~c3 3 3π²0 ~c3

The 2p → 1s transition in hydrogen has an energy of (3/4)RH = 10.2 eV, and therefore ω = 1.55 × 1016 rad/s. Hence we obtain A2p→1s = 6.27 × 108 s−1 . We then see from eqn 5.2 that the radiative lifetime τR = 1/A = 1.6 ns. This agrees with the experimental value. (5.4) It follows from eqn 5.4 that: 1 1 1 = + , τ τR τNR where 1/τNR is the non-radiative recombination rate. Hence the excited state lifetime can be shorter than the radiative lifetime. The radiative lifetime τR would normally be independent of T , since it depends only on the electron wave functions. However, the non-radiative recombination rate 1/τNR generally increases with T due to increased non-radiative recombination by phonon emission or thermally-activated traps. Hence τ decreases with T . The quantum efficiency can calculated from eqn 5.5 to be equal to τ /τR . At 300 K we find ηR = 79 %, while at 350 K we find ηR = 56 %. (5.5) Semiconductors emit light at their band gap energy. We are therefore looking for a semiconductor with Eg = 2.3 eV. Inspection of Table C.3 suggest two possibilities: ZnTe or GaP. The latter has an indirect band gap and would therefore not be very efficient. Hence the most likely candidate material is ZnTe. Note, however, that we could also make an emitter at this wavelength by using an alloy, for example: Inx Ga1−x N. (See Fig. 5.11.) Linear extrapolation between the GaN and InN band gaps would suggest that a composition with x ≈ 0.41 would emit at 540 nm. (5.6) (a) Consider an incremental beam slice at a position z within the absorbing material as shown in Fig. 16. Let A be the area of the slice, dz its thickness, I the incoming intensity (power per unit area), and δI the loss of intensity due 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 time is therefore equal to AδI/hν. Hence the carrier generation rate G per unit volume is given by: G=

AαIdz/hν αI AδI/hν = = . Adz Adz hν 46


Name: Tapas Banerji


IP: 129.82.195.78

(b) The rate equation for the carrier density N is dN N Iα N =G− = − , dt τ hν τ where the first term accounts for carrier generation and the second for carrier recombination. In steady-state conditions we must have that dN = 0. dt Hence: N=

Iατ . hν

(c) We calculate the laser intensity from: I=

P 1 mW = = 1.3 × 105 Wm−2 . A π (50 µm)2

The sample is antireflection coated, and so this is also the intensity inside the sample. We can then calculate the carrier density using the result from part (b) with the values of τ and α given in the exercise. Hence we find N = 6.6 × 1020 m−3 for hν = 2.41 eV. Note that this is the carrier density 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. dz Incoming intensity I(z) Incoming photon number N (z)

Outgoing intensity I - dI Outgoing photon number N-dN

A z

Figure 16: An incremental slice of a laser beam at a position z within an absorbing material. A is the area of the slice and dz its thickness. In Exercise 5.6 we consider a continuous laser beam with an intensity I(z), whereas in Exercise 5.7 we consider a pulse with a photon number of N(z). (5.7) (a) The argument proceeds along similar lines as for the previous exercise. Consider again an incremental beam slice of area A and thickness dz, as shown in Fig. 16. Since we are now dealing with a pulse rather than a continuous beam, we need to consider the photon number rather than the intensity. Let N be the incoming photon number, and δN the number of photons 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 recombination takes place while the material is being excited. Hence the number 47


Name: Tapas Banerji


IP: 129.82.195.78

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

δN αNdz αN = = . Adz Adz A

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

αE . Ahν

On inserting the relevant numbers from the exercise, we find N = 1.9 × 1024 m−3 . (b) The pulse excites the carrier density calculated in part (a) at time t = 0. The carriers then recombine and the carrier density decreases according to N (t) = N0 exp(−t/τ ) , where τ is the total decay rate including both radiative and non-radiative recombination. (See eqn 5.4.) With τ = (1/τR + 1/τNR )−1 = 0.89 ns, we find N (t) = N0 /2 at t = 0.62 ns. (c) The quantum efficiency is calculated from eqn 5.5 as ηR = 89%. Each laser pulse contains E/hν photons. We are told that the crystal is ‘thick’, and so we can assume that all the laser photons are absorbed in the crystal. The number of photons re-emitted by luminescence is therefore EηR /hν = 3.5 × 1010 photons.

E

Ee Eg

hn

Eh k=0

k

Figure 17: Definition of energies as required for Exercise 5.8. (5.8) The emission rate is proportional to the probability that the upper level is occupied and that the lower level is empty. These probabilities can be 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 hole occupancy of the lower level (i.e. the probability that the lower level is empty.) Hence the emission probability is proportional to fe × fh . 48


Name: Tapas Banerji


IP: 129.82.195.78

In the classical limit, the occupancy factors follow Boltzmann statistics with: fe,h ∝ exp(−Ee,h /kB T ) , where Ee and Eh are the electron and hole energies within the conduction or valence band, respectively. (See Fig. 17.) Hence: fe fh ∝ exp(−Ee /kB T ) · exp(−Eh /kB T ) = exp[−(Ee + Eh )/kB T ] . Now it is apparent from Fig. 17 that hν = Eg + Ee + Eh , and hence that (Ee + Eh ) = hν − Eg . We therefore conclude that: I(hν) ∝ fe fh ∝ exp[−(hν − Eg )/kB T ] . (5.9) The number of electrons in the conduction band is given by eqn 5.6. In the classical limit, we have (E − EF ) À kB T , so that the electron occupancy factor is given by: fe (E) =

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

On using the density of states given in eqn 5.7, we then obtain: 1 Ne = 2π 2

µ

2m∗e ~2

¶3/2

µ exp

EF kB T

¶Z

µ

∞

1/2

(E − Eg ) Eg

exp

−E kB T

¶ dE .

On introducing the variable x = (E − Eg )/kB T we then obtain: 1 Ne = 2π 2

µ

2m∗e kB T ~2

¶3/2

µ exp

(EF − Eg ) kB T

¶Z

∞

x1/2 exp(−x) dx .

0

The final result is obtained by setting (EF − Eg ) ≡ EFc , where EFc is the electron Fermi energy measured relative to the bottom of the conduction band. For the case of GaAs at 300 K, we can insert the effective mass and temperature, and use the definite integral given in the Exercise to obtain: µ c ¶ EF × (4.4 × 1023 ) m−3 . Ne = exp kB T In part (a) we then find EFc = −0.216 eV ≡ −8.4kB T , whereas in part (b) we find EFc = +0.021 eV ≡ +0.83kB T . The approximations are therefore valid for part (a) because we have (E − EF ) À kB T for all states in the conduction band, but not for part (b), where the Fermi level comes out above the conduction band minimum. The aim of this exercise is to get the students to think about the conditions under which the use of Boltzmann statistics is valid.

49


Name: Tapas Banerji


IP: 129.82.195.78

(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, and replace the Fermi function by unity up to this energy. With the energies measured relative to the band edge, we then find: µ ∗ ¶3/2 Z EFc 2me 1 E 1/2 dE , Ne = 2π 2 ~2 0 for the electrons and similarly for the holes. On evaluating the integral we find: µ ∗ ¶3/2 1 2 2me Ne = × (EFc )3/2 , 2π 2 ~2 3 and similarly for the holes. Equation 5.13 then follows by simple rearrangement. (5.11) We use eqn 5.13 to evaluate the Fermi energies. In part (a) we find EFc = 0.36 meV for the electrons, and EFv = 0.073 meV for the holes. In part (b) we find EFc = 36 meV for the electrons, and EFv = 7.3 meV for the holes. For the conditions of degeneracy to apply, we need EF À kB T . In part (a), the electrons will be degenerate for T ¿ 4.2 K, while the holes will be degenerate for T ¿ 0.9 K. In part (b), the electrons will be degenerate for T ¿ 420 K, while the holes will be degenerate for T ¿ 85 K. This shows that it is much easier to obtain degenerate statistics for the electrons than the holes because they are lighter. (5.12) The Fermi wave vector kF is, in general, related to the Fermi energy EF by: ~2 kF2 EF = . 2m With EF given by eqn 5.13, we then find: kF = (3π 2 N )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 . Hence the Fermi wave vectors of the electrons and holes are identical. The fact that the mass does not appear in the formula for the Fermi wave vector is a consequence of the fact that the density of states in k-space does not depend on the mass. (See eqn 3.15.) (5.13) (a) The solid angle subtended by an object of area dA at a distance r is given by (See Fig 18(a)): dA dΩ = 2 . r The lens will be positioned at a distance equal to its focal length from the sample, and so we set r = 100 mm in this case. We then find: dΩ =

π(25 mm/2)2 = 0.049 . (100 mm)2 50


Name: Tapas Banerji


IP: 129.82.195.78

dA

qinside

dW

qoutside

S

r

medium n (a)

air n=1

(b)

Figure 18: (a) Definition of solid angle dΩ for an object of area dA at a distance of r, as required for Exercise 5.13. (b) Emission of a ray at angle θinside from a source S embedded within a medium of refractive index n. (b) Consider a ray emitted by a source embedded within a medium of refractive index n as illustrated in Fig. 18(b). The ray is refracted at the surface according to Snell’s law, with: sin θoutside = n. sin θinside The light is being collected by a lens of radius 12.5 mm positioned 100 mm from the surface, and therefore only those rays that satisfy θoutside ≤ 12.5/100 will be collected (assuming small angles). By Snell’s law we deduce that only those rays within a cone with θinside ≤ 0.125/n will be collected. 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 is worked out by considering a circle of radius rθinside placed at a distance of r from the source: F =

dA πr2 (θinside )2 (θinside )2 (θoutside )2 = = ≈ . 4πr2 4πr2 4 4n2

Finally, we have to consider that some of the photons will be reflected at the surface. The reflectivity is calculated from eqn 1.29 to be 21% for n = 2.7. Hence the final fraction collected is 0.79F = 4.2 × 10−4 . (c) The semiconductor absorbs photons of energy 2.41 eV and emits luminescent photons at the band gap energy of 1.61 eV with a probability equal to ηR . Hence the power emitted is equal (1.61/2.41)ηR times the power absorbed. The incoming laser will be partially reflected at the surface, and so the maximum power that can be absorbed is equal to (1 − R) × Pincident = 0.79 × 1 mW. Hence the maximum luminescent power is equal to: Plum =

1.61 ηR × 0.79 × 1 mW = 0.53 ηR mW . 2.41

(d) We obtain the luminescent power collected by the lens by multiplying 51


Name: Tapas Banerji


IP: 129.82.195.78

the total luminescent power by the collected fraction: P collected = 0.53 ηR × 4.2 × 10−4 mW = 0.22 ηR µW . (5.14) (a) The electron Fermi energy is calculated from eqn 5.13 to be 0.14 eV. (b) The hole Fermi energy is calculated from: Z Nh

=

v EF

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

0

µ ¶3/2 2 1 3/2 3/2 (mhh + mlh ) E 1/2 dE 2 2 2π ~ 0 µ ¶3/2 1 2 3/2 3/2 (mhh + mlh ) (EFv )3/2 . 2 3π ~2

Z = =

v EF

Hence for Nh = 2 × 1024 m−3 we obtain EFv = 0.012 eV. (c) The carriers will be degenerate if EF > kB T . At 180 K we have kB T = 0.018 eV, and so the electrons are degenerate, but not the holes. (d) When both the electrons and holes are degenerate, we expect emission from the band edge up to the Fermi energies as illustrated in Fig. 5.7. We then expect emission from the band gap to Eg + EFc + EFv . At T = 0 we would have a sharp cut-off at this energy, but at finite T the edge is broadened over ∼ kB T . The luminescence would then be expected to fall to 50 % of its peak value at Eg + EFc + EFv . On inserting the calculated Fermi energies, and the value of the band gap, we expect the 50% point at around 0.95 eV. This can be compared to the experimental value of ∼ 0.94 eV. The agreement between the experiment and model is thus good. (e) The luminescence spectrum at 250 ps given in Fig. 5.8 is consistent with a value of the electron Fermi energy of ∼ 0.035 eV. This implies from eqn 5.13 that the carrier density is Ne ≈ 3 × 1023 m−3 . The electrons are still degenerate at this density for T = 55 K. The lifetime is estimated from: N (t) = N0 exp(−t/τ ) , which implies N (24)/N (250) = exp(−226/τ ), where τ is the lifetime in ps. On inserting the two values, we find τ ≈ 120 ps = 0.12 ns. (5.15) We consider a zinc–blende III–V semiconductor with the band structure shown in Fig. 3.8. As explained in Section 3.3.7, the heavy-hole transitions are three times as strong as the light-hole transitions. The selection rules are ∆MJ = ∓1 for σ ± light, as indicated in Fig. 19. Note that these selection rules are the opposite way around to those for absorption, because we are now considering emission. Consider the situation after excitation with σ + light resonant with the band gap. There will be three times as many electrons in the MJ = −1/2 state as the +1/2 state. On the other hand, the holes will be equally distributed between all four possible sub-levels due to fast spin–orbit mixing. The transitions that are possible are indicated in Fig. 19. The intensity 52


Name: Tapas Banerji


IP: 129.82.195.78

E Eg

conduction band

3N s+

hh

-D

J = 1/2

s+

3 0

N

1

3

1

s-

lh

slh

so

so

hh

J = 3/2

J = 1/2

valence band -3/2

-1/2

+1/2

+3/2

MJ

Figure 19: Optical transitions after excitation with σ + photons, as required for Exercise 5.15. The relative weights of the transitions are indicated. will be proportional to the population of the initial electron state and the transition weight (i.e. |M |2 ). From the MJ = −1/2 electron state with population 3N we have two transitions: • A σ + transition to the MJ = −3/2 hh level: relative intensity = 3 × 3 = 9. • A σ − transition to the MJ = +1/2 lh level: relative intensity = 3 × 1 = 3. From the MJ = +1/2 electron state with population N we also have two transitions: • A σ + transition to the MJ = −1/2 lh level: relative intensity = 1 × 1 = 1. • A σ − transition to the MJ = +3/2 hh level: relative intensity = 1 × 3 = 3. The total relative σ + intensity I + is therefore 9 + 1 = 10, while the total relative σ − intensity I − is 3 + 3 = 6. The polarization is then given by eqn 5.14 as: 10 − 6 4 I+ − I− = = = 25% . P = + I + I− 10 + 6 16 The argument works equally well for σ − excitation. (5.16) We see from eqn 5.16 that P (B)/P (0) = 1/2 when ΩTS = 1. Hence at B1/2 we have (from eqn 5.17): ΩTS =

ge µB B1/2 TS = 1. ~

53


Name: Tapas Banerji


IP: 129.82.195.78

Hence (cf eqn 5.18): ge µB B1/2 1 1 1 ≡ + = . TS τ τS ~ Now from eqn 5.15 we have that: P (0) 1 = , P0 1 + τ /τS which implies:

1 1 1 P0 = + . τ P (0) τ τS

Hence:

ge µB B1/2 1 P0 = , τ P (0) ~

which gives: τ=

P0 ~ . ge µB B1/2 P (0)

We can then work out τS from: ge µB B1/2 1 1 = − , τS ~ τ which implies: τS =

~ ge µB B1/2

µ ¶−1 P (0) 1− . P0

Note that P0 is expected to be 25% for a bulk III-V semiconductor, as shown in Exercise 5.15. (5.17) We follow Example 5.1. We extrapolate linearly between the direct gaps of GaP and GaAs to find that: Egdirect (x) = 1.42 + 1.36x eV . (a) The LED will cease to function efficiently when the gap becomes indirect, i.e. for x > 0.45, where the conduction band minimum at the Brillouin zone edge drops below the conduction band minimum at k = 0. Hence the minimum wavelength occurs for x = 0.45, where Eg = 2.03 eV. This corresponds to a wavelength of 610 nm. (b) We require a direct band gap equivalent to 670 nm, namely 1.85 eV. On substituting into the formula for the direct gap, we find x = 0.316. Note that this is less than 45%, so that the alloy will have a direct gap and hence be an efficient emitter. (5.18) (a) The reflectivity is calculated from eqn 1.29 to be 31% for n = 3.5. (b) The cavity mode frequency is given by eqn 5.20, and implies a mode spacing of ∆ν = c/2nl. With n = 3.5 and l = 1 mm, we find ∆ν = 4.3 × 1010 Hz.

54


Name: Tapas Banerji


IP: 129.82.195.78

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

1 ln(0.95 × 0.31) = 610 m−1 . 2 × 10−3 m

(5.19) (a) The maximum possible power would be obtained if one photon is emitted for each electron that flows through the device. The power is then equal to the electron flow rate multiplied by the photon energy: µ ¶ 100 mA Pmax = hν × = 150 mW . e (b) The electrical power input is equal to IV = 0.1 × 1.9 = 0.19 W = 190 mW. With a power output of 50 mW, the power conversion efficiency is therefore equal to (50/190) = 26 %. (c) The power output will vary as shown in Fig. 5.15(a). The slope efficiency is calculated from eqn 5.26 as (50/(100 − 35) = 0.77 mW/mA = 0.77 W/A. The differential quantum efficiency is calculated from eqn 5.26 to be 51 %. (5.20) The incident primary electron flux per unit area is equal to J/e electrons m−2 . Each primary electron will generate N eh electron hole pairs in a distance Re from the surface, where N eh is given by eqn 5.27 and Re is the penetration depth. The generation rate per unit volume of crystal is thus given by: J J Ep G= × N eh = (1 − γ) i . e Re e Re E The rate equation for the number of electron–hole pairs is thus given by: dN N =G− , dt τ where τ is the lifetime. In steady state conditions we must have N˙ = 0 (cf Exercise 5.6), and we thus have: N = Gτ =

Jτ Ep (1 − γ) i . e Re E

55


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 6 Quantum confinement (6.1) Substitute into eqn 6.3 with ∆x = 10−6 m and m = 0.1m0 to obtain T . 0.01 K. (6.2) The energy difference between the n = 1 and n = 2 levels is worked out from eqn 6.13 to be 3~2 π 2 /2m∗ d2 . On setting this energy difference to be equal to kB T /2, we then derive the required result, namely s 3~2 π 2 d= . m∗ kB T On substituting into this formula with T = 300 K, we find d = 9.3 nm for m∗ = m0 , and d = 30 nm for m∗ = 0.1m0 . √ On comparing with eqn 6.4, it is apparent that d = 3π 2 ∆x. This shows that the two criteria used to determine when quantum size effects are important give the same answer, apart from a numerical factor of ∼ 5. The differing numerical factor is to be expected, given the approximate nature of the criteria.

ky dk

k kx

(2p/L)2

Figure 20: Grid of allowed k states in a two-dimensional material of area L2 , as required for the solution of Exercise 6.3. (6.3) The x and y components of the k vector must each satisfy the criterion exp(ikL) = 1, which implies k = integer × 2π/L. The possible values of the k vector therefore form a regular grid in k-space as shown in Fig. 20, with a grid-spacing of 2π/L. The area per k-state is (2π/L)2 , and the 56


Name: Tapas Banerji


IP: 129.82.195.78

density of states for an area L2 is therefore L2 /(2π)2 . This implies that the density of states in k-space for a unit area of crystal is 1/(2π)2 . It is also apparent from Fig. 20 that the area of k-space enclosed by the increment k → k + dk is 2πkdk. If we call the number of k-states enclosed by this area g L (k)dk, we then find: g L (k)dk =

2πkdk L2 kdk . = 2 (2π/L) 2π

Hence for a unit area of crystal we have: g2D (k)dk =

k dk . 2π

The density of states in energy space can be found from eqns 3.13–14. (The factor or two accounts for the fact that spin 1/2 particles have two spin states for each k-state.) With E(k) = ~2 k 2 /2m, we have dE/dk = ~2 k/m, and hence 2g2D (k) 2k/2π m g2D (E) = = 2 = , dE/dk ~ k/m π~2 as required. (6.4) In one dimension the k states must all lie on a single axis (say the x axis). Let L be the length of the wire. The Born–von Karman boundary conditions require that exp(ikL) = 1, and hence that k = integer × 2π/L. The density of states in k space for a sample of unit length is therefore 1/2π. We thus have: 1 g1D (k)dk = dk . 2π The density of states in energy space is then found from eqns 3.13–14 with E = ~2 k 2 /2m. This gives: g1D (E) =

³ m ´1/2 2g1D (k) 2/2π m = 2 = = E −1/2 , dE/dk ~ k/m π~2 k 2~2 π 2

as required. (6.5) The depth of the potential well enters eqn 6.26 via ξ, which is defined in eqn 6.27. The √ function on the right hand√side of eqn 6.26 decreases from (m∗w /m∗b )1/2 ξ at x = 0 to zero at x = ξ. (See, for example, Fig. 6.5, for the case with ξ = 13.2.) The function on the RHS of eqn 6.26 will therefore always cross the x tan x function between 0 and π/2, no matter how small ξ is. (6.6) We follow the method of Example 6.1. We have µ

and ξ=

m∗w m∗b

¶1/2

µ =

0.34 0.5

¶1/2 = 0.82 ,

0.34m0 × (10−8 )2 × 0.15 eV = 33.5 . 2~2 57



IP: 129.82.195.78

y = x tan x

6

4

y = 0.82 (33.5-x2)1/2 x = 1.30

y

Name: Tapas Banerji

2

0

0

2

x

4

6

Figure 21: Graphical solution required for Exercise 6.6. Hence we must solve: x tan x = 0.82

p 33.5 − x2 .

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

2~2 x2 = 7.6 meV , m∗w d2

for d = 10 nm. The equivalent energy for an infinite well is calculated from eqn 6.13 to be 11 meV. (6.7) (a) The wave functions of an infinite potential well form a complete orthonormal basis, with Z +∞ ϕ∗n ϕn0 dz = δnn0 , −∞

where δnn0 is the Kronecker delta function: δnn0

= =

1 if n = n0 0 if n 6= n0 .

It is apparent from eqn 6.11 that the wave functions depend only on n and are therefore identical for electrons and holes with the same n. Hence the orthonormality condition applies irrespective of whether ϕn and ϕn0 are electron or hole wave functions, or a mixture. Hence: Mnn0

= 1 if n = n0 = 0 if n = 6 n0 .

This result can also be derived by explicit (and somewhat tedious) substitution of the wave functions from eqn 6.11 into the formula for Mnn0 . (b) This result follows from parity arguments. The wave functions of a finite well have well-defined parities as a consequence of the inversion 58



IP: 129.82.195.78

symmetry about the centre of the well. Wave functions with odd n have even parity, while those with even n have odd parity. Hence the product ϕ∗en ϕ∗hn0 has even parity if n − n0 is even and has odd parity for odd n − n0 . The integral of an odd function from −∞ → +∞ is zero, and so the overlap integral is zero if ∆n is equal to an odd number. n 1 2

Electron 225 898

Heavy hole 30 120

Light hole 188 750

Table 1: Confinement energies in meV calculated for an infinite quantum well of width 5 nm, as required for Exercise 6.8. (6.8) The confinement energies of the electrons, heavy holes and light holes calculated from eqn 6.13 are given in Table 1. With an infinite well, we only need consider ∆n = 0 transitions. The threshold photon energies for these transitions are given by eqn 6.39, or the equivalent for higher values of n or for the light holes. Two transitions fall in the range 1.4 → 2.0 eV: • Heavy hole 1 → electron 1 at 1.679 eV, • Light hole 1 → electron 1 at 1.837 eV. For each transition we expect a step at the threshold energy as in Fig. 6.9. In 2-D materials the joint density of states is proportional to the electronhole reduced mass µ (see eqn 6.41), and hence the relative height of the heavy-hole and light-hole transition steps is in proportion to their reduced masses, that is 0.059 : 0.036. Hence the final spectrum would appear as in Fig. 22. 1.837 eV Absorption (arb. units)

Name: Tapas Banerji

0.1 1.679 eV

lh1 ® e1

hh1 ® e1 0.0 1.4

1.6 1.8 Photon energy (eV)

2.0

Figure 22: Absorption spectrum for Exercise 6.8 (6.9) (a) In finite wells the confinement energies are reduced compared to those of an infinite well of the same width. Hence the transition energies would be lower. Furthermore, transitions that are forbidden in infinite wells become weakly allowed, such as the hh3 → e1 transition. This transition would fall within the observed energy range, as might also some n = 2 transitions.

59


Name: Tapas Banerji


IP: 129.82.195.78

(b) Peaks would appear below the steps in the absorption spectrum due to excitonic absorption. The difference in energy between the peak and the continuum absorption would be equal to the exciton binding energy. RR ∗ ψ ψ dA = 1. For the (6.10) (a) To prove normalization, we must show that given wave function we have: µ ¶ Z ∞ Z 2π Z ∞ Z 2π 2 ∗ Ψ Ψ rdrdφ = exp(−2r/ξ) rdrdφ , πξ 2 r=0 φ=0 r=0 φ=0 ¶ µ Z ∞ 2 × 2π × exp(−2r/ξ) rdr , = πξ 2 r=0 = 1, as required. ˆ on Ψ, using the fact that ∂Ψ/∂φ = 0: (b) We first compute the effect of H µ ¶ ~2 d dΨ e2 ˆ HΨ = − r − Ψ, 2µr dr dr 4π²0 ²r r µ 2 ¶ ~ e2 Ψ ~2 Ψ + − . = − 2µξ 2 2µξ 2 4π²0 ²r r We can then calculate hEivar from: Z hEivar

∞

Z

2π

= r=0 Z ∞

ˆ rdrdφ , Ψ∗ HΨ

φ=0 Z 2π

µ 2 ¶ ¸ · ~ e2 Ψ ~2 Ψ + − rdrdφ , Ψ∗ − 2 2 2µξ 2µξ 4π²0 ²r r r=0 φ=0 µ 2 ¶ Z ∞ Z 2π Z ∞ Z 2π ~2 ~ e2 ∗ = − Ψ Ψ rdrdφ + − Ψ∗ Ψ drdφ , 2µξ 2 r=0 φ=0 2µξ 2 4π²0 ²r r=0 φ=0 µ 2 ¶ ~2 ~ e2 2 = − ×1+ − × , 2 2 2µξ 2µξ 4π²0 ²r ξ e2 ~2 = + − . 2µξ 2 2π²0 ²r ξ =

(c) On differentiating hEivar with respect to ξ, we find that hEivar achieves its minimum value of Emin = −µe4 /8(π²0 ²r ~)2 for ξ = 2π~2 ²0 ²r /µe2 . The minimum energy is four times larger than the bulk exciton binding energy found in Exercise 4.4. (d) In part (c) we found ξmin = 2π~2 ²0 ²r /µe2 . This can be written in terms of the bulk exciton radius aX defined in eqn 4.2 as: ξmin = aX /2 . Hence the radius of the exciton in 2-D is half the radius of the bulk. (6.11) At d = ∞ we have bulk GaAs, while at d = 0 we have bulk AlGaAs. For intermediate values of d, we have a GaAs quantum well exciton with an enhanced binding energy. In an ideal 2–D system we would expect 60



IP: 129.82.195.78

four times the binding energy of bulk GaAs (i.e. 16 meV), but in realistic systems, the enhancement might be smaller due to the imperfect quantum confinement of the finite-height AlGaAs barriers. Thus as d is reduced from ∞, 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 might typically be around 10 meV.

hh1 ® e1 band edge 1.0 hh1 ® e1 exciton

PLE intensity

Name: Tapas Banerji

0.0

lh1 ® e1 exciton lh1 ® e1 band edge

1.58

1.60 1.62 Energy (eV)

1.64

Figure 23: Interpretation of the data in Fig. 6.23 as required for Exercise 6.12. (6.12) (a) See Section 5.3.5. (b) The interpretation of the principal features in the data is shown in Fig. 23. For both heavy- and light-hole transitions, we expect to observe a peak due to excitonic absorption followed by a step at the band edge. The two strong peaks observed in the data correspond to the excitons for the hh1 → e1 and lh1 → e1 transitions, while the flat absorption bands above both excitons correspond to the interband transitions. These interband absorption bands are flat because the density of states is independent of the energy in 2-D systems (see eqn 6.41.) (c) The hh1 → e1 interband continuum starts at 1.592 eV. In the infinite well 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 well model overestimates the confinement energy. (d) The binding energies can be read from Fig. 6.23 as the energy gap between the exciton peak and the appropriate band edge. With the transitions identified as in Fig. 23, we find binding energies of 11 meV for the heavy holes and 12 meV for light holes. For a perfect 2-D system we would expect 4 × RX for GaAs, i.e. 16.8 meV. The experimental values are lower because a real quantum well is not a perfect 2-D system. (6.13) By considering Fig. 6.12 we would expect three different regimes to apply. 1. For photon energies in the range set by eqn 6.43, that is: Eg + Ee1 + Ehh1 ≤ ~ω < Eg + Ee1 + Elh1 , we can only excite hh1 → e1 transitions. These only create MJ = −1/2 spin down electrons, and so we expect Πe = −100%. 61


Name: Tapas Banerji


IP: 129.82.195.78

2. For photon energies in the range: Eg + Ee1 + Elh1 ≤ ~ω < Eg + Ee1 + Eso1 , where Eso1 is the confinement energy of the first split-off hole band, we can excite both hh1 → e1 and lh1 → e1 transitions, which create oppositely polarized electrons. Hence we expect the magnitude of the electron spin polarization to decrease. If the matrix elements are the same as in the bulk, the heavy-hole transitions will be three times stronger than the light-hole transitions, and this gives an electron spin polarization of −50%. (See Section 3.3.7.) 3. For photon energies in the range: ~ω > Eg + Ee1 + Eso1 , we expect that no spin polarization will be be created. The extra contribution of the split-off hole band exactly cancels the net polarization created by the stronger weight of the heavy-hole transition. This can be understood in general terms by realizing that the electron spin polarization is created by spin–orbit coupling, so that when the excess photon energy exceeds the spin–orbit interaction energy ∆, then the effect will be negligible. Experimental data showing these effects may be found in Vi˜ na, J. Phys.: Condens. Matter 11, 5929 (1999), Fig. 2. The basic trends discussed above are reproduced in the experimental data, although there are some important differences. This is because we really should work in the excitonic rather than single-particle picture, and also because the relative weight of the heavy- and light-hole transitions is not exactly 3:1. See the discussion in Vi˜ na’s article for further details. (6.14) (a) With H 0 = +ezE z , we have: Z ∆E (1) = eE z

+∞

ϕ∗ zϕ dz .

−∞

Since z is an odd function, and ϕ∗ ϕ ≡ |ϕ|2 is an even one, the integral is zero. (b) With H 0 = +ezE z , the second-order energy shift is given by: ∆E (2) = e2 E 2z

∞ X |h1|z|ni|2 . E1 − En n=2

Since the wave functions have parity (−1)n+1 , and z is an odd parity operator, all the terms with odd n are zero. Hence we have: ∆E (2) = e2 E 2z

|h1|z|4i|2 |h1|z|2i|2 + e2 E 2z + ··· . E1 − E2 E1 − E4

Now the terms with n ≥ 4 are much smaller than the term with n = 2. (This can be verified by working through the integrals, but it is fairly 62


Name: Tapas Banerji


IP: 129.82.195.78

obvious given the larger denominator.) Hence we only need to consider the first term: |h1|z|2i|2 . ∆E (2) = e2 E 2z E1 − E2 On substituting the energies for an infinite well from eqn 6.13, this becomes: 2e2 E 2z m∗ d2 |h1|z|2i|2 . ∆E (2) = − 3~2 π 2 Now, on redefining the origin so that the quantum well runs from z = 0 to z = +d, we have: Z +∞ h1|z|2i = ϕ∗1 z ϕ2 dz , −∞ Z d

= =

2 d

−

sin(πz/d) z sin(2πz/d) dz , 0

16d . 9π 2

Hence we find ∆E

(2)

¯ ¯2 µ ¶6 2 2 ∗ 4 2e2 E 2z m∗ d2 ¯¯ 16d ¯¯ e Ezm d 2 =− × ¯− 2 ¯ = −24 , 3~2 π 2 9π 3π ~2

as required. (6.15) (a) Figure 6.13 predicts that the energy of the hh1→e1 transition will shift from 1462.4 meV to 1438.5 meV in a 10 nm quantum well on increasing the field from zero to 1 × 107 V/m. This corresponds to a red shift of 14.1 nm. This can be compared to the experimental red shift of 10.5 nm for a 9 nm well for a voltage change of 10 V. Excitonic effects are not included in the calculation of Fig. 6.13, but this is not expected to make a large difference. Two main factors account for the difference between the experimental and theoretical results. • The well width for the experiment was smaller by a factor of 0.9. From eqn 6.47 we would then expect the Stark shift to be smaller by a factor (0.9)4 = 0.66. • For the experimental data, we can calculate the field from eqn 6.52. From this we find that E z changes from 0.15 × 107 V/m to 1.15 × 107 V/m, which should be compared to the theoretical data, which corresponds to changing E z from 0 → 1 × 107 V/m. The field change is the same in both cases, but because the Stark shift is quadratic at low fields, we do not expect the Stark shifts to be the same. If the Stark shift remained quadratic at all field strengths, we would expect the experiment shift to be larger by a factor (1.152 −0.152 )/(12 −02 ) = 1.30. Putting these two factors together, we expect the experimental Stark shift to be smaller than the theoretical one by a factor of 0.94 × 1.3 = 0.85. The actual ratio is (10.5/14.1) = 0.74. This difference is easily explained by

63


Name: Tapas Banerji


IP: 129.82.195.78

the fact that the Stark shift does not remain quadratic at all fields, and so the factor due to the different fields should be smaller. (b) We assume that the Stark shift is quadratic, and hence that ∆E ∝ E 2z . We work out the field strength from eqn 6.52, which implies E z = 11.5 MV/m at 10 V and E z = 6.5 MV/m at 5 V. For small shifts we have ∆λ ∝ ∆E, and so we find: µ ∆λ(5 V) =

6.5 11.5

¶2 ∆λ(10 V) = 0.32 × 10.5 nm = 3.4 nm .

(c) The wavelength red shift of 10.5 nm corresponds to an energy shift of –18 meV. This energy shift is related to the net electron-hole displacement h∆zi by: ∆E = −pz E z = −eh∆ziE z . With E z = 11.5 MV/m, we thus obtain h∆zi = 1.6 nm. E z (MV/m) 3 6 9

Sample A ∆Ecalc ∆Eexp 1.5 1 5.9 5 13 13

Sample B ∆Ecalc ∆Eexp 15 6 62 27 139 54

Table 2: Comparison of the calculated and measured Stark shifts (in meV) for the two samples discussed in Exercise 6.16. (6.16) The experimental data is taken from Polland et al., Phys. Rev. Lett. 55, 2610 (1985). We analyse it by using the energy shift calculated by secondorder perturbation theory in Exercise 6.14(b). Since we are considering an electron-hole transition, we must add together the Stark shifts of the electrons and holes, giving: µ ∆E

(2)

= −24

2 3π

¶6

e2 E 2z d4 ∗ (me + m∗hh ) . ~2

The calculated shifts using m∗e = 0.067m0 and m∗hh = 0.5m0 are compared to the experimental ones in Table 2. It is apparent that the model works well for sample A, but not for sample B. The model breaks down when the size of the Stark shift becomes comparable to the energy splitting of the unperturbed hh1 and hh2 levels. This is essentially the same criterion as for the transition from the quadratic to the linear Stark effect in atomic physics. In sample B, we are in this regime at all the fields considered. (6.17) At E z = 0 the quantum well is symmetric about the centre of the well. The electron and hole states therefore have a definite parity with respect to inversion about z = 0. The parity is (−1)(n+1) , and the electron–hole overlap given by eqn 6.36 is zero if ∆n is odd: see Exercise 6.7(b). At finite E z , the inversion symmetry is broken, and the states no longer have a definite parity. Therefore, the selection rule based on parity no longer holds.

64


Name: Tapas Banerji


IP: 129.82.195.78

(6.18) With the confinement energy E ∝ d−2 , we have dE/dd ∝ −2/d3 , and hence we expect: ∆E ∆d = −2 . E d A ±5% change in d is thus expected to give ∆E/E = ±10%. With E = 0.1 eV, we then expect a full-width broadening of 0.02 eV. This is comparable to the linewidth observed in the 10 K data shown in Fig. 6.16. A ±5% variation in d corresponds more or less to a fluctuation of one atomic layer. Such “monolayer” fluctuations are unavoidable in the crystal growth. The linewidth at room temperature is further broadened by the thermal spread of the carriers in the bands. (6.19) We assume infinite barriers and use eqn 6.53 to calculate the emission energy. 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 narrower to compensate for the imperfect confinement of the barriers. (6.20) (a) z is an odd function with respect to inversion about z = 0. The integral from −∞ to +∞ will therefore be zero unless ϕ∗n ϕn0 is also an odd function, which requires that the wave functions must have different parities. Since the wave functions have parities of (−1)n+1 , the condition is satisfied if n is an even number and n0 odd, and vice versa. Hence ∆n must be equal to an odd number. (b) The strength of the intersubband transitions is proportional to the square of the matrix elements. These matrix elements can be evaluated by substituting the wave functions from eqn 6.11. On redefining the origin so that the quantum well runs from z = 0 to z = +d, we find: 2 h1|z|2i = d and

2 h1|z|4i = d

Z

d

sin(πz/d) z sin(2πz/d) dz = − 0

Z

d

sin(πz/d) z sin(4πz/d) dz = − 0

16 d, 9π 2

4 d. 45π 2

Hence the 1 → 4 transition is weaker than the 1 → 2 transition by a factor [(4/45)/(16/9)]2 = [1/20]2 = 2.5 × 10−3 . It is apparent from Fig. 6.17 that the wavelength of the 1 → 2 transition is given by hc 3π 2 ~2 = E2 − E1 = , λ 2m∗e d2 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, and hence λ = 29 µm. (6.21) Consider a ray incident at angle θ to the normal as shown in Fig. 24. The ray will be refracted according to Snell’s law, with sin θ = n, sin θ0

65


Name: Tapas Banerji


air n=1

E

semiconductor, refractive index n

IP: 129.82.195.78

z

q

q¢ E¢

quantum well

Figure 24: Refraction of light entering a semiconductor containing a quantum well, as discussed in Exercise 6.21. where θ0 is the angle inside the crystal. For intersubband transitions, we are interested in the z component of the electric field of the light at the quantum well, namely: E z = E 0 sin θ0 = E 0 sin θ/n . If I0 is the incident intensity, and there are no intensity losses at the surface, then the intensity in the z component at the quantum well is given by Iz = I0 (sin θ/n)2 , since the intensity is proportional to E 2 . Hence the fraction of the power of the beam in the z polarization at the quantum well is (sin θ/n)2 . This fraction has a maximum value of 1/n2 for θ = 90◦ . Therefore even if we completely absorb all the z polarized light by intersubband transitions, and we use glancing incidence, we can only remove a fraction of 1/n2 of the power in the incident beam. This fractional absorption is equal to 9% if n = 3.3. nx 1 2 2 3 2 3 3

ny 1 1 2 1 2 2 2

nx 1 1 1 1 2 1 2

(n2x + n2y + n2z ) 3 6 9 11 12 14 17

g 1 3 3 3 1 6 3

Table 3: Quantum numbers of the energy levels for a cubic quantum dots in order of increasing energy, as discussed in Exercise 6.22. g denotes the degeneracy excluding spin. (6.22) For a cubic dot, the energies of the quantized levels are given by eqn 6.54 with dx = dy = dz = d, implying: E(nx , ny , nz ) =

π 2 ~2 (n2 + n2y + n2z ) , 2m∗ d2 x

where nx , ny and nz are integers with a minimum value of 1. As demonstrated by Table 3, the quantized levels occur at energies of 3, 6, 9, 11, 12, 14, 17,. . . in units of h2 /8m∗ d2 . 66


Name: Tapas Banerji


IP: 129.82.195.78

The degeneracy factor g is worked out by finding the number of permutations of (nx , ny , nz ) that can give the same energy. The ground state is unique, but for the first excited we can have (2, 1, 1), (1, 2, 1) and (1, 1, 2), and similarly for the levels at 9, 11, and 17 h2 /8m∗ d2 . The level at 14 h2 /8m∗ d2 has a degeneracy of 6 because there are 3! = 6 ways of arranging the numbers 3, 2 and 1. (6.23) The radial equation for a spherical dot is given by eqn 6.57. If the confining potential is V0 , then we can put V (r) = −V0 for r ≤ R0 . Therefore, when the particle is inside the dot (i.e. r ≤ R0 ), the radial wave function for states with l = 0 must satisfy: · µ ¶ ¸ ~2 1 d d − ∗ 2 r2 − V0 R(r) = E R(r) . 2m r dr dr Now

1 d r2 dr

µ ¶µ ¶ µ ¶ sin kr sin kr 2 d 2 r = −k . dr r r

Hence if R(r) = sin kr/r, then substitution into the radial equation gives: ~2 k 2 R(r) − V0 R(r) = E R(r) , 2m∗ which implies that

~2 k 2 . 2m∗ For an infinite well it must be the case the R(r) = 0 for r ≥ R0 . Therefore at r = R0 we require that sin kR0 /R0 = 0, which implies that kR0 = nπ, where n is an integer. Hence: E = −V0 +

E = −V0 +

~ 2 n2 π 2 . 2m∗ R02

The confinement energy relative to the bottom of the potential well at −V0 is thus: ~2 n2 π 2 En = . 2m∗ R02 (6.24) For the dots to have the same volume, we require that: d3 =

4 3 πR , 3 0

where d is the cube size and R0 is the radius of the spherical dot. This implies that (R0 /d) = (3/4π)1/3 . From Exercise 6.22 we find that the confinement energy of the first level in a cubic dot with infinite barriers is E1cubic = 3~2 π 2 /2m∗ d2 . For a spherical dot eqn 6.58 gives E1spherical = ~2 π 2 /2m∗ R02 for C10 = 1, as appropriate for infinite barriers. Hence: E1cubic E1spherical

=

~2 π 2 3~2 π 2 ÷ =3 ∗ 2 2m d 2m∗ R02

µ

R0 d

¶2

µ =3

3 4π

¶2/3 = 1.15 .

67


Name: Tapas Banerji


IP: 129.82.195.78

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

Figure 25: Comparison of cube and sphere of the same volume, as discussed in Exercise 6.24. The wave functions have nodes at the edges. (6.25) (a) In Cartesian co-ordinates, the Schrödinger equation for a 2–D harmonic oscillator takes the form: µ 2 ¶ ¸ · ∂ ∂2 1 ~2 2 2 2 + 2 + me ω0 (x + y ) Ψ(x, y) = EΨ(x, y) . − 2me ∂x2 ∂y 2 On writing Ψ(x, y) = X(x)Y (y) and substituting, we find: −

~2 d2 X 1 ~2 d2 Y 1 Y + me ω02 x2 XY − X 2 + me ω02 y 2 XY = EXY . 2 2me dx 2 2me dy 2

On dividing by XY and re-arranging this gives: −

~2 1 d2 Y ~2 1 d2 X 1 1 2 2 x = E + + m ω − me ω02 y 2 . e 0 2me X dx2 2 2me Y dy 2 2

The left hand side is a function of x, and the right hand side a function of y, and so they both must equal a constant. On calling this constant C and rearranging we have: 1 ~2 d2 X + me ω02 x2 X 2 2me dx 2 ~2 d2 Y 1 − + me ω02 y 2 Y 2me dy 2 2

−

= CX = (E − C)Y .

Both of these are standard harmonic oscillator equations. The one for x implies C = (nx + 1/2)~ω0 , where nx is an integer, while the one for y implies that (E − C) = (ny + 1/2)~ω0 , where ny is another integer. Hence: E = C + (ny + 1/2)~ω0 = (nx + ny + 1)~ω0 ≡ (n + 1)~ω0 , where n = nx + ny . The degeneracies are found by working out the different possible permutations of nx and ny to get the same energy: 68


Name: Tapas Banerji


IP: 129.82.195.78

• Ground state: E = ~ω0 , n = 0, (nx , ny ) = (0, 0), degeneracy = 1. • 1st excited state: E = 2~ω0 , n = 1, (nx , ny ) = (1, 0) or (0,1), degeneracy = 2. • 2nd excited state: E = 3~ω0 , n = 2, (nx , ny ) = (2, 0), (1,1) or (0,2), degeneracy = 3. • 3rd excited state: E = 4~ω0 , n = 3, (nx , ny ) = (3, 0), (2,1), (1,2) or (0,3), degeneracy = 4. • 4th excited state: E = 5~ω0 , n = 4, (nx , ny ) = (4, 0), (3,1), (2,2), (1,3) or (0,4), degeneracy = 5. It is thus apparent that the degeneracy is equal to n. This results can also be worked out by considering the m states: there are n states from −n to +n in steps of 2. (b) In polar co-ordinates, the Schrödinger equation for a 2–D harmonic oscillator takes the form: · µ ¶ ¸ ~2 1 ∂ ∂ 1 ∂2 1 2 2 − r + 2 2 + me ω0 r ψ(r, φ) = Eψ(r, φ) . 2me r ∂r ∂r r ∂φ 2 On writing ψ(r, φ) = R(r)Φ(φ) and substituting, we find: µ ¶ ~2 1 d dR R d2 Φ 1 − r Φ+ 2 + me ω02 r2 RΦ = ERΦ . 2me r dr dr r dφ2 2 On multiplying through by r2 /RΦ and re-arranging, this gives: µ ¶ ~2 r d dR 1 1 d2 Φ − r + me ω02 r4 − Er2 = − . 2me R dr dr 2 Φ dφ2 The left hand side is a function of r, and the right hand side a function of φ, and so they both must equal a constant. On calling this constant C and rearranging we have: d2 Φ = −CΦ , dφ2 which has solutions Φ(φ) = eimφ , where m2 = C. Now the wave function must be single valued for each value of φ and so we require: Φ(φ) = Φ(φ + 2π) , which implies that eim2π = 1 and hence that m must be an integer. Therefore, the wave functions are of the form ψ(r, φ) = R(r)eimφ , where m is an integer. (c) The fact that m is equal to −n to +n in steps of 2 can be proved mathematically, but requires a lot of work. (See, for example, Pauling & Wilson, Introduction to quantum mechanics, 1935.) We thus restrict our efforts here to justifying this result for the first three levels. From part (a) we know that the wave function for the state with energy (n+1)~ω has the form ψnx (x)ψny (y), where n = nx +ny . We can draw up a correspondence 69


Name: Tapas Banerji


Energy ~ω 2~ω 3~ω

Polar co-ordinates n = 0, m = 0 n = 1, m = ±1 n = 2, m = −2, 0, +2

IP: 129.82.195.78

Cartesian co-ordinates (nx , ny ) = (0, 0) (nx , ny ) = (1, 0) or (0, 1) (nx , ny ) = (2, 0), (1, 1) or (0, 2)

Table 4: Equivalence of the states of a 2–D harmonic oscillator in polar and Cartesian co-ordinates, as considered in Exercise 6.25(c). between the Cartesian and polar co-ordinates as shown in Table 4. We consider each energy state in turn. Ground state with energy ~ω This state is easy. There is only one possibility with (nx , ny ) = (0, 0). The wave function in polar co-ordinates is therefore of the form: ψ0,0 (r, φ) = ψ0 (x)ψ0 (y) ∝ exp(−αx2 /2) · exp(−αy 2 /2) = exp(−αr2 /2) , as required. First excited state with energy 2~ω We now have two possibilities, and the wave functions must be linear combinations of each other, that is: ψ1,m (r, φ) = C1 ψ1 (x)ψ0 (y) + C2 ψ0 (x)ψ1 (y) . It is apparent that if we take ψ1,m (r, φ) ∝ ψ1 (x)ψ0 (y) ± iψ0 (x)ψ1 (y) , then we have: ψ1,m (r, φ) ∝ (x ± iy) exp(−αx2 /2) · exp(−αy 2 /2) . Now x = r cos φ and y = r sin φ. Therefore: ψ1,m (r, φ) ∝ r(cos φ ± i sin φ) exp(−αr2 /2) = r e±iφ exp(−αr2 /2) . This shows that m = ±1 for the state with energy 2~ω. Note that the radial form is as given in the Exercise. Second excited state with energy 3~ω We now have three possibilities, and the wave functions must be linear combinations of each other, that is: ψ2,m (r, φ) = C1 ψ2 (x)ψ0 (y) + C2 ψ1 (x)ψ1 (y) + C3 ψ0 (x)ψ2 (y) . Consider the combination ψ2 (x)ψ0 (y) + ψ0 (x)ψ2 (y). This can be written: ψ2,m (r, φ) ∝

(1 − 2αx2 ) exp(−αx2 /2) · exp(−αy 2 /2)

∝

+ exp(−αx2 /2) · (1 − 2αy 2 ) exp(−αy 2 /2) , 2(1 − αx2 − αy 2 ) exp(−α(x2 + y 2 )/2) ,

∝

2(1 − αr2 ) exp(−αr2 /2) . 70


Name: Tapas Banerji


IP: 129.82.195.78

This clearly has no dependence on φ and so it must correspond to the state with m = 0. Note that the radial form is correct for this state. We can thus write: ψ2,0 (r, φ) ∝ ψ2 (x)ψ0 (y) + ψ0 (x)ψ2 (y) . √ Now consider the combination: −ψ2 (x)ψ0 (y)± 2iψ1 (x)ψ1 (y)+ψ0 (x)ψ2 (y). By carefully considering the normalization constants in front of the harmonic oscillator functions, we find that this gives: ψ2,m (r, φ) ∝ ∝ ∝ ∝ ∝ ∝

(−1 + 2αx2 ± 4iαxy + (1 − 2αy 2 )) exp(−α(x2 + y 2 )/2) , 2α(x2 − y 2 ± 2ixy) exp(−αr2 /2) , (r2 cos2 φ − r2 sin2 φ ± 2ir2 cos φ sin φ) exp(−αr2 /2) , r2 ((cos2 φ − sin2 φ) ± 2i cos φ sin φ) exp(−αr2 /2) , r2 (cos 2φ ± i sin 2φ) exp(−αr2 /2) , r2 e±2iφ exp(−αr2 /2) .

By comparing with the radial functions given in the Exercise, we then conclude that √ ψ2,2 (r, φ) ∝ −ψ2 (x)ψ0 (y) + 2iψ1 (x)ψ1 (y) + ψ0 (x)ψ2 (y) , √ ψ2,−2 (r, φ) ∝ −ψ2 (x)ψ0 (y) − 2iψ1 (x)ψ1 (y) + ψ0 (x)ψ2 (y) . This shows that the allowed values of m for n = 2 are indeed −2, 0 and +2. It can be verified that these three wave functions are orthogonal, i.e. R that they satisfy ψ ∗ ψ d2 r = 0.

71


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 7 Free electrons (7.1) The method for determining the electron Fermi energy was considered in Exercise 5.10, and the formula for EF is given in eqn 5.13: EF =

~2 (3π 2 N )2/3 , 2m

which implies:

µ ¶3/2 1 2mEF . 3π 2 ~2 On substituting this form of N into the formula for the plasma frequency in eqn 7.6, we then obtain µ ¶4 9²2 ~2 π~ωp EF3 = 0 . 8m e N=

(7.2) We expect 100% reflectivity below the plasma frequency and transmission at higher frequencies. (See Fig. 7.1.) Hence we can set ωp /2π = 3 MHz in this example. On substituting into eqn 7.6 and solving for N , we find N ∼ 1011 m−3 . (n.b. The electron density calculated here is just a typical one. 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 the value of σ for salt water we find δ ∼ 0.5 m for ω/2π = 200 kHz. This shows that electromagnetic waves of this frequency penetrate less than 1 m from the surface of the sea. To obtain a skin depth of 30 m as required for communication with a submarine submerged at this depth, we require ω/2π = 70 Hz. Even lower frequencies are required for deeper depths. The data 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 transmission edge at 440 nm implies λp = 440 nm, and hence: ωp = 2πc/λp = 4.3 × 1015 rad/s . On substituting into eqn 7.6 and solving for m, we find m∗e = 1.4 × 10−30 kg ≡ 1.6 m0 . (7.5) This Exercise closely follows Example 7.3. We read ωp = 1.36 × 1016 rad/s from Table 7.1, and work out τ = 4.0 × 10−14 s from eqn 7.14 using the value of N from Table 7.1 and the value of σ0 given in the exercise. At 500 nm we have ω = 3.77 × 1015 rad/s, and the relative permittivity at this wavelength is then given by eqns 7.16–17 as: ²˜r = −12.0 + 0.086i . We finally use eqns 1.25–26 to calculate n ˜ = −0.012 + 3.5i, and substitute n = −0.012 and κ = 3.5 into eqn 1.29 to obtain R = 99.6 %. 72


Name: Tapas Banerji


IP: 129.82.195.78

(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 extinction coefficient 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 ²˜r = −52.1 + 0.235i, and hence κ = 7.22 (see eqn 1.26). This value of κ implies, through eqn 1.19, an absorption coefficient α = 9.1 × 107 m−1 . The transmission is given by Beer’s law (see eqn 1.4) as exp(−αz) = 0.16 for z = 20 nm. Note that the skin depth approximation in eqns 7.18–20 is not valid here because we have ωτ = 230 for λ = 1 µm (ω = 1.9 × 1015 rad/s), and hence we do not satisfy ωτ ¿ 1. The absorption coefficient estimated from eqn 7.19 is wrong by approximately one order of magnitude. (7.7) Based on the plasma frequency of gold given in Table 7.1, we expect high reflectivity above ∼ 140 nm. The low reflectivity up to 600 nm is caused by interband transitions as illustrated schematically in Fig 7.4. The energy gap between the d bands and the Fermi energy can be read from the data as the energy equivalent of ∼ 520 nm, i.e. ∼ 2.4 eV. It is apparent from the reflectivity spectrum that gold reflects red, orange and yellow light stronger than green and blue. This accounts for its characteristic yellowish colour. (7.8) It is apparent from eqn 1.29 that R = 0 when n ˜ = 1, and hence ²r = 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 zero reflectivity when: (see eqn 7.24 for ωp ): ! Ã ωp2 N e2 1 = ²opt − ∗ = ²opt 1 − 2 . m ²0 ω 2 ω Equation 7.25 is then derived by solving this formula for ω 2 . N (1024 m−3 )

λ(R = 0) (µm)

ω(R = 0) (1013 rad/s)

m∗e /m0

0.35 0.62 1.2 2.8 4.0

33 27 22 16 14

5.7 7.0 8.6 12 13

0.020 0.028 0.036 0.044 0.048

Table 5: Effective masses of n-type InSb calculated from the data in Fig 7.7 as required for Exercise 7.9. (7.9) The easiest way to determine the effective mass from the spectra is to read the wavelength at which R = 0 from the data. We then use eqns 7.24–25 to write: N e2 , m∗ = ²0 (²opt − 1)ω 2 where ω is the angular frequency corresponding to R = 0. The values of the effective masses found in this way with ²opt = 15.6 are given in Table 5. 73


Name: Tapas Banerji


IP: 129.82.195.78

It is apparent from Table 5 that m∗e increases strongly with N , rising from 0.020 m0 at 3.5 × 1023 m−3 to 0.048 m0 at 4 × 1024 m−3 . This increase in the effective mass with carrier density is caused by non-parabolicity in the conduction band of InSb. The effective mass approximation assumes that the bands are parabolic in shape as in Fig 3.5. However, a glance at the 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 good fit 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 +

~2 k 2 , 2m∗ (k)

where m∗ (k) = m∗e for small k, but then increases at larger values of k. As we dope the semiconductor with more electrons, the Fermi energy increases (see eqn 5.13) and we are probing states of the conduction band with larger values of E and k. It is therefore to be expected that the effective mass should increase with the doping density. The band non-parabolicity is particularly strong in some narrow gap semiconductors such as InSb. (7.10) The free carrier absorption coefficient in the limit ωτ À 1 is given by eqn 7.28. At 10 µm we have ω = 1.9 × 1014 rad/s, and on inserting the relevant 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 valence bands which can then induce free carrier absorption. The carrier density generated 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 carrier lifetime, and ~ω is the photon energy. On inserting the appropriate values as given in the Exercise, we find N = 1.9 × 1022 m−3 . This is the density of electrons in the conduction band and holes in the valence band, both of which cause free carrier absorption. At λ = 10.6 µm we have ω = 1.8 × 1014 rad/s, and hence ωτ = 36 for the electrons and ωτ = 9 for the holes. We thus have ωτ À 1 in both cases, and we can therefore calculate the free carrier absorption coefficients by using eqn 7.28. This gives α = 130 m−1 for the electrons and α = 70 m−1 for the holes. Hence the total free carrier absorption 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 of τ . In the calculation of the carrier density, τ represents the carrier lifetime, whereas in eqn 7.28 it represents the momentum scattering time. • Intervalence band absorption has been neglected here. However, with only ∼ 1022 m−3 holes, it is probable that intervalence band absorption will be insignificant. (See the next exercise.) 74


Name: Tapas Banerji


IP: 129.82.195.78

• This exercise is actually an example of a nonlinear optical effect: the light beam at 633 nm induces changes in the optical properties. The mechanism is that the beam creates carriers which then alter the optical properties. These types of nonlinear effects are called “free carrier nonlinearities” for obvious reasons. See Chapter 11 for more information on nonlinear optics.

E max,min

min

k

EF

max

min

D

max

lh

hh

kF

kF

Figure 26: Intervalence band transitions as required for the solution of Exercise 7.12. The Fermi energy is not drawn to scale: with the parameters given in the Exercise, the Fermi energy is just above the split-off band. The labels (1), (2) and (3) refer to the lh→hh, SO→lh, and SO→hh transitions as in Fig. 7.9. (7.12) (a) The holes are distributed between the heavy- and light-hole bands, and so we can find the Fermi energy from (see eqn 3.16): Z N

=

EF

(ghh + glh ) dE 0

=

23/2 3/2 3/2 (m + mlh ) 2π 2 ~3 hh

Z

EF

E 1/2 dE .

0

With m∗hh = 0.5 m0 and m∗lh = 0.08 m0 we then find EF = 0.032 eV for N = 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. The wave vector at the Fermi energy is worked out from: EF =

~2 kF2 . 2m∗

This give kFhh = 6.5 × 108 m−1 and kFlh = 2.6 × 108 m−1 for the heavy and light holes respectively. (See Fig. 26.)

75


Name: Tapas Banerji


IP: 129.82.195.78

(b) The energies of the three types of intervalence band transitions indicated in Fig. 7.9 can be calculated by using eqns 3.18–20: µ ¶ 1 ~2 1 − k2 , (1) lh → hh : ~ω = |Elh (k) − Ehh (k)| = 2 m∗lh m∗hh µ ¶ ~2 1 1 (2) SO → lh : ~ω = |Eso (k) − Elh (k)| = ∆ + − k2 , 2 m∗so m∗lh µ ¶ 1 ~2 1 − ∗ k2 , (3) SO → hh : ~ω = |Eso (k) − Ehh (k)| = ∆ + 2 m∗so mhh where k is the wave vector at which the transition occurs. Transitions can only take place from occupied states to empty ones. The upper and lower limits of the transition energies are therefore set by the Fermi wave vectors as indicated in Fig. 26. The energy limits calculated using the Fermi wave vectors from part (a) are therefore as follows: 1. lh→hh transitions: The lower and upper limits are set by kFlh and kFhh 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 than the split-off mass, the lower and upper limits correspond to the transitions at kFlh and k = 0 respectively. On inserting into (2) above we find 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 kFhh , giving a range of 0.34 → 0.42 eV. (7.13) (a) The energies of the 2p0 , 3p0 and 4p0 transitions can be read from Fig. 7.11 as 34.0, 40.1, and 42.4 meV respectively. These energies fit well to the formula: µ ¶ 1 ∗ hν = R 1 − 2 , n with R∗ = 45.2 meV. This is consistent with eqn 7.30 if we set m∗e = 0.85 m0 for the donor levels. (b) The energies of the 2p± , 3p± , 4p± and 5p± transitions can be read from Fig. 7.11 as 39.2, 42.4, 43.3 and 44.1 meV respectively. The np± transitions correspond to transitions from np hydrogenic states to the 1s state, with energies given by: hν = |E1s − Enp± | . The 1s energy is just equal to the value of R∗ found in the part (a), and ∗ so we can identify R0∗ = R∗ = 45.2 meV. The value of R± is then found ∗ by fitting the transition energies. A good fit is found with R± ≈ 25 meV. (7.14) These are donor level transitions as shown in Fig. 7.10(a). Their transition energies are given by eqn 7.30. It is apparent that they depend only on the effective mass and relative permittivity of the host crystal, and not on any of the properties of the dopant atom. The reason why this is so is that the donor levels are formed by the interaction between a band 76


Name: Tapas Banerji


IP: 129.82.195.78

electron and an ionized impurity. The donor atoms are chosen so that they have a single excess electron, and so they are all singly ionized. The donor level energies are therefore hydrogenic with the relevant mass and permittivity of the band electron. On comparing with eqn 7.30, we see that we must have R∗ = (m∗e /m0 ²2r )× RH . On setting this equal to 2.1 meV we find m∗e = 0.036 m0 for ²r = 15.2. conduction band

Eg

acceptor levels valence band

Figure 27: Acceptor to conduction band transitions in a p-type semiconductor as discussed in Exercise 7.15. Note that this figure is not drawn to scale: the acceptor energies are much smaller than the band gap. (7.15) Doping with acceptors creates a p-type semiconductor with a series of acceptor levels just above the valence band as indicated in Fig. 27. The energy of the acceptor levels relative to the top of the valence band will be given by eqn 7.29 with m∗e replaced by m∗h . Electrons from the valence band can be thermally excited to fill the acceptor levels, giving rise to the possibility of acceptor to conduction band transitions as indicated in Fig. 27. These transitions occur at a photon energy of Eg − EA . The absorption edge of a p-type semiconductor will therefore decrease on doping from Eg to Eg − EA , where EA is the energy of n = 1 acceptor level. A wavelength shift from 5.26 µm to 5.44 µm corresponds to an energy shift of 8 meV, and hence we deduce EA ∼ 8 meV. This exercise is based on experimental data for zinc acceptors in InSb. See Figure 2 in “Impurity and Exciton Effects on the Infrared Absorption Edges of III-V Compounds”, E.J. Johnson H.Y. Fan, Phys. Rev. 139, A1991–A2001 (1965). (7.16) The peak is caused by Raman scattering from plasmon modes, as shown in Fig. 7.13. The energy of the scattered photons is given by eqn 7.49. In this exercise, we are clearly considering the case of plasmon emission where the ‘–’ sign is appropriate. We thus deduce: ~ωp = ~ωin − ~ωout = (2.410 − 2.321) eV = 0.089 eV . On substituting into eqn 7.24 with ²opt = n2 and the given value of m∗ , we find N = 4.2 × 1024 m−3 . Note that we would also expect Raman signals from optical phonons, but these would occur at smaller energy shifts with ∆ν ∼ 300 cm−1 ≡ 37 meV. (See Fig. 10.11.) 77


Name: Tapas Banerji


(7.17) We set

µ ωp =

N e2 ²opt ²0 m∗e

IP: 129.82.195.78

¶1/2 = ΩLO ,

with ²opt = n2 and solve for N . A wave number ν of 297 cm−1 implies ω = 2πcν = 5.6 × 1013 rad/s, and hence we find N = 7.2 × 1023 m−3 . A mixed plasmon–phonon mode is formed at this doping density because the two longitudinal excitations couple strongly together if their frequencies are the same. (7.18) We first calculate ωp using eqn 7.6. With N = 1 × 1029 m−3 , we find ωp = 1.8 × 1016 rad/s. Next, we work out vF from: EF =

1 mv 2 . 2 F

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

~ (3π 2 N )1/3 , m

which gives vF = 1.7 × 106 m/s. A wave vector of 0.1π/a has a magnitude of 7.9 × 108 m−1 if a = 4 ˚ A. We then have: µ ¶ 3 (1.7 × 106 )2 (7.9 × 108 )2 ω ≈ ωp 1 + = 1.002 ωp . 10 (1.8 × 1016 )2 We thus see that the term in k 2 only makes a difference of 0.2%. It can therefore safely be ignored in most circumstances. (7.19) In an electron energy loss experiment on a metal, we expect to observe peaks from both the surface and bulk plasmons. On inserting ²d = 1 into eqn 7.61 as appropriate for √ air, we expect the two plasmons to differ in frequency by a factor of 2 = 1.41. The data fits this model quite well, with the experimental ratio being 1.49. The 10.3 eV and 15.3 eV series are thus the surface and bulk plasmons respectively. The bulk plasma energy of 15.3 eV implies ωp /2π = 3.7 × 1015 Hz. It is then apparent from Table 7.1 that the metal is aluminium. This Exercise is based on the electron energy loss spectroscopy data shown in Kittel (7th edition), Chapter 10, Figure 8. (7.20) (a) At this point, we ignore the imaginary part of ²m , and put ²d = 1 and ²m = −18 into eqn 7.62. With ω/c = 2π/λ, this gives: kzd

=

2π 600 × 10−9

kzm

=

2π 600 × 10−9

µ µ

−12 −18 + 1

¶1/2

−(−18)2 −18 + 1

= 2.5 × 106 m−1 , ¶1/2 = 4.6 × 107 m−1 .

Hence lzd = (kzd )−1 = 394 nm and lzm = (kzm )−1 = 22 nm. 78


Name: Tapas Banerji


IP: 129.82.195.78

(b) The propagation length in the x direction is determined by the imaginary part of kx and so in this part of the exercise we do have to include the imaginary part of ²m . On substituting ²d = 1 and ²m = (−18 + i) into eqn 7.58 with ω/c = 2π/λ, we find: 2π kx = 600 × 10−9

µ

−18 + i −17 + i

¶1/2 = 1.05 × 107 (1.03 + 1.7 × 10−3 i) m−1 .

On writing the real and imaginary parts of kx as k 0 and k 00 respectively, we then obtain k 00 = 1.8 × 104 m−1 . The wave therefore propagates with: E(x) = E 0 exp[ikx x] = E 0 exp[(i(k 0 + ik 00 )x] = E 0 exp(ik 0 x) exp(−k 00 x) . The intensity is proportional to EE ∗ , and therefore decays with x according to I(x) = I0 exp (−2k 00 x). The propagation distance is therefore given by (2k 00 )−1 = 28 µm. (7.21) Equation 7.63 shows that the resonance in the polarizability of a colloidal metal nanoparticle occurs when ²m = −2²d . On setting ²d = n2 for the dielectric, and using the standard frequency dependence of ²m for a metal (cf. eqn 7.7), we see that this resonance occurs when: 1−

ωp2 = −2n2 . ω2

On rearranging, this gives: (1 + 2n2 )ω 2 = ωp2 , and hence: ω=√

ωp . 1 + 2n2

√ When the dielectric is air, we have n = 1, and hence ω = ωp / 3. (7.22) (a) As shown in the discussion of eqn 7.63, the resonance occurs when ²m = −2²d . When ²m is complex, as is the case considered here, this resonance condition is satisfied when the real part of ²˜m , namely ²1 , is equal to −2²d . This is because ²d is real, and so the denominator of eqn 7.63 is smallest when ²1 + 2²d = 0. For water we have ²d = n2 = 1.77, and so we require ²1 = −3.5. A linear fit to the data given in Table 7.3 indicates that this condition is satisfied at about 517 nm. (b) The result derived in Exercise 7.21 predicts a resonance wavelength given by: √ p 2πc 1 + 2n2 λ= = = 1 + 2n2 λp , ω ωp /2π where λp is the wavelength corresponding to ωp . On using the value of λp for gold given in Table 7.1, namely 138 nm, we predict λ = 294 nm, which differs significantly from the correct value obtained in part (a). The reason for the discrepancy is that the undamped Drude formula given in eqn 7.7 is not a good approximation for gold in the optical frequency range. For example, the Drude model would predict high reflectivity up to 79


Name: Tapas Banerji


IP: 129.82.195.78

the plasma wavelength of 138 nm, but this is not observed in experimental reflectivity data. (See Fig. 7.20.) At optical frequencies the d-band transitions are very strong, and this modifies the relative permittivity very significantly. In fact the situation is more complicated. Equation 7.63 only applies to spherical nanoparticles. Furthermore, it predicts that the resonance frequency should be independent of the nanoparticle size, which is not actually the case: see Maier (2007) for details. This becomes apparent when attempting to fit the data in Fig. 7.17. With ²d = 2.5, we would expect the resonance to occur when ²1 = −5. Table 7.3 gives ²1 = −5 at 538 nm for gold, but the peak actually occurs at 580 nm. This shows that a more detailed model is required to get a good fit to the data. Equation 7.63 does, however, explain the basic phenomenon, and correctly explains the red shift of the resonance with increasing ²d .

80


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 8 Molecular materials (8.1) This is a standard example covered in all elementary quantum mechanics texts. On substituting Ψ1 into the Schrödinger equation, we obtain: µ ¶ ~2 x2 1 1 − − Ψ1 + mΩ2 x2 Ψ1 = E1 Ψ1 , 4 2 2m a a 2 which can be re-arranged to give: ¶ µ ~2 ~2 1 2 2 mΩ − x Ψ + Ψ1 = E1 Ψ1 . 1 2 2ma4 2ma2 Ψ1 is therefore a solution if we set a = (~/mΩ)1/2 to eliminate the first term. We can then deduce: E1 =

~2 1 = ~Ω . 2ma2 2

By a similar method we can show that Ψ2 and Ψ3 are solutions with E2 = (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 of width d are given by eqn 6.13 as: En =

h2 n 2 . 8md2

Selection rules permit transitions with ∆n equal to an odd number. (See the discussion of intersubband transitions in Section 6.7.) The lowest energy transition occurs for n = 1 → 2, with hν = 3h2 /8md2 . On setting m = m0 for the π-electron, and hν = 2.5 eV, we find d = 6.7 × 10−10 m. This corresponds to about six carbon–carbon bonds. Note that this model should not be taken too seriously, since it predicts that the transition energy scales as d−2 . The data in Fig. 8.11 show that the actual scaling is much weaker, with E ∼ d−0.4 . (8.3) The ratio of the number of molecules with one vibrational quantum excited compared to those with none is given by Boltzmann’s law as exp(−hν/kB T ), where ν is the frequency of the vibration. The calculated ratios for the three modes at 300 K are: • ν = 2 × 1013 Hz: 4 × 10−2 , • ν = 4 × 1013 Hz: 1.6 × 10−3 , • ν = 7 × 1013 Hz: 1.4 × 10−5 . 81



IP: 129.82.195.78

The point of this exercise is to make the student appreciate that it is a good approximation to assume that all the molecules are in the vibrational ground state at room temperature. (8.4) The energy of isolated hydrogen atoms is given by the standard Rydberg formula with En = −RH /n2 . The energy required to promote one of the atoms from a 1s to a 2p state is therefore (3/4)RH = 10.2 eV. This is smaller than the equivalent transition in the H2 molecule. The reason is that the energy of a diatomic molecule is given by: E molecule = E atom1 + E atom2 − E binding , where E atomi is the energy of the isolated atoms, and E binding is the binding energy of the molecule. Now the ground state of the molecule is more strongly bound than the excited state, and the transition includes the difference of the two binding energies. Hence in the molecule the transition energy is: hν = (3/4)RH + ∆E binding . We can then account for the molecular transition energy of 11.3 eV if we assume that the difference of the binding energies of the 1s1s and 1s2p molecular configurations is 1.1 eV. 1.0 0.8 0.6 U(r)

Name: Tapas Banerji

0.4 0.2 0.0

r 0.5

1.0

1.5

2.0

-0.2

Figure 28: Lennard–Jones potential for A = B = 1, as considered in Exercise 8.5. (8.5) (a) The attractive r−6 term is caused by the van der Waals interaction, which is the main attractive force between neutral molecules. This is a dipole-dipole interaction. A fluctuating dipole p1 on molecule 1 generates an electric field of strength E 1 ∝ p1 /r3 at molecule 2. This induces a dipole of magnitude p2 ∝ E 1 ∝ p1 /r3 on molecule 2, which then generates a field E 2 ∝ p2 /r3 ∝ p1 /r6 at molecule 1. The interaction energy is then −p1 E 2 ∝ −(p1 )2 /r6 . Although the time average of p1 will be zero, the time average of (p1 )2 is not. Hence the dipole-dipole mechanism generates an attractive potential ∝ r−6 . (b) The Lennard-Jones potential is plotted for the case A = B = 1 in Fig. 28. The potential is attractive at large r, but the repulsive term dominates for small r. This gives a minimum at a well-defined value of r, 82


Name: Tapas Banerji


IP: 129.82.195.78

which we call r0 . The position of the minimum energy can be calculated by setting: dU 12A 6B = − 13 + 7 = 0 , dr r r at r = r0 , which implies r0 = (2A/B)1/6 . (c) The Taylor expansion for U (r) near r0 is: µ ¶ µ ¶ dU 1 d2 U U (r) = U (r0 ) + (r − r0 ) + (r − r0 )2 + · · · , dr 2 dr2 where the derivatives are evaluated at r = r0 . If r0 is the position of the minimum, then the first derivative will be zero, and U(r) reduces to: µ ¶ 1 d2 U U (r) = U (r0 ) + (r − r0 )2 + · · · . 2 dr2 Now

d2 U 156A 42B = + 14 − 8 , 2 dr r r

which, with r = r0 = (2A/B)1/6 , gives: 18B 2 d2 U = 2 dr A

µ

B 2A

¶1/3 .

Thus: U (r) = ≡

µ ¶1/3 18B 2 B (r − r0 )2 + · · · , A 2A 1 U (r0 ) + µΩ2 (r − r0 )2 . 2

U (r0 ) +

Hence Ω2 = (18B 2 /Aµ)(B/2A)1/3 . (8.6) (a) This result follows directly from the Franck–Condon principle. We are at T = 0 and so the initial state of the system will be the lowest vibrational level of the lower state, i.e. the n = 0 vibrational level of the electronic ground state. The initial wave function is therefore ϕ0 (Q − Q0 ). Transitions are possible to any of the vibrational levels of the excited state, as shown in Fig. 8.7. The energy of the transition to the nth level is given by eqn 8.3 and can be written: ~ωn = ~ω0 + n~Ω . We are at T = 0 and thus expect negligible broadening of the transitions. The probability for the transition is given by the Franck–Condon factor (eqn 8.12), which in this case takes the form: ¯Z ¯ P (n, 0) = ¯¯

∞ 0

ϕ∗n (Q

−

Q00 )

¯2 ¯ ϕ0 (Q − Q0 ) dQ¯¯ ≡ |hn, Q00 |0, Q0 i|2 .

83



IP: 129.82.195.78

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

I(~ω) ∝

n=0 ∞ X

∝

P (n, 0)δ(~ω − ~ωn ) |hn, Q00 |0, Q0 i|2 δ(~ω − ~ω0 − n~Ω) .

n=0

as required.

S=0

Intenmsity (arb units)

Name: Tapas Banerji

S=1

S=5

0

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

Figure 29: Vibronic spectra, as considered in Exercise 8.6. (b) We substitute for the Franck–Condon factor to obtain: I(~ω) ∝

∞ X exp(−S)S n δ(~ω − ~ω0 − n~Ω) . n! n=0

This defines a Poisson distribution. The spectra are shown in Fig.29. (i) If S = 0, then the only non-zero term is the one with n = 0. The spectrum therefore just consists of the zero-phonon line at ~ω0 . (ii) When S = 1, the strongest two transitions are the ones with n = 0 and n = 1. The intensity drops sharply with increasing n. (iii) We now have a bell-shaped distribution peaked √ at n = 5. The half width of the distribution is given approximately by 2 5 ∼ 5. In the limit of large S, the spectrum is Gaussian. More details about how the Huang–Rhys factor determines the vibronic line shape may be found, for example, in Henderson & Imbusch, Optical spectroscopy of inorganic solids (Oxford, 1989) or Hayes & Stoneham, Defects and defect processes in nonmetallic solids (Wiley, 1985). 84



Transition energy (eV)

IP: 129.82.195.78

5.4 5.2 5.0 4.8 4.6 0

2

4

6

Quantum number n Figure 30: Analysis of transition energies of benzene as considered in Exercise 8.7. (8.7) The absorption spectrum will consist of a series of vibronic lines with energies given by eqn 8.3. The longest wavelength will correspond to the transition with n = 0, and the others to increasing values of n. The transition energies are plotted against this assignment of n in Fig. 30. The good linear fit confirms the assignment. The fitting parameters that come out of the analysis are ~ω0 = 4.65 eV and ~Ω2 = 0.113 eV. On making the reasonable assumption that ~|Ω2 − Ω1 | ¿ (E2 − E1 ), we then deduce from eqn 8.4 that E2 = E1 +4.65 eV. We thus identify the energy of the S1 state relative to the S0 ground state as 4.65 eV, and find Ω/2π = 2.7 × 1013 Hz.

E 0.13 eV

S2 0.11 eV n=6

S1

7.3 eV

n=5

5.7 eV

Name: Tapas Banerji

S0

Q

Q0

Figure 31: Configuration diagram for ammonia deduced from the data of Fig. 8.9 as considered in Exercise 8.8. (8.8) The absorption spectrum shows a progression of vibronic lines obeying eqn 8.3 with at least two excited electronic states. The first progression 85


Name: Tapas Banerji


IP: 129.82.195.78

starts at ∼ 5.7 eV, and has a vibrational splitting of 0.11 eV. The intensity of the lines peaks for n = 6. The second progression starts at ∼ 7.3 eV, has a 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 of Q0 for the states. The intensity peaks when the overlap of the vibronic states is largest. The wave function of the ground state S0 peaks at Q0 , while for the excited states the wave function gradually peaks more and more 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 align respectively with Q0 for S0 . We thus obtain the schematic configuration diagram given in Fig. 31. (8.9) The spin–orbit interaction introduces a coupling mechanism between the spin and orbital angular momenta S and L. It is therefore possible to have an interaction between different S states via their spin–orbit interaction with a common L state. This interaction produces a small mixing of the wave functions, so that triplet states contain a small admixture of singlet character. This small singlet admixture gives a finite probability for a triplet-to-singlet transition, which would otherwise be totally forbidden if the spin states were pure. Spin-orbit coupling is now routinely used to increase the intensity of phosphorescence in organic LEDs. A heavy metal (eg platinum) is introduced into the molecule, and this increases the spin–orbit interaction, because the spin–orbit coupling generally scales as Z 2 , where Z is the atomic number. The use of heavy-metal dopants strongly enhances the triplet-to-singlet transition rate, and hence the overall light emission efficiency. This is especially important in electrically-pumped devices, which are otherwise limited to a maximum efficiency of 25%. (See Exercises 8.16 and 8.17.) (8.10) The weak nature of the emission and the long radiative lifetime indicates that we are dealing with phosphorescence from a triplet state. The energy level scheme for pyrromethene 567 would be qualitatively similar to that for anthracene shown in Fig. 8.12(b). From the spectra shown in Fig. 8.10 we deduce that the S1 level has an energy of 2.3 eV, while the wavelength of the phosphorescence indicates that the triplet lies at an energy of 1.6 eV. In the case of optical excitation, the phosphorescence could be caused by intersystem crossing from excited singlet states. (8.11) The assignment of the vibronic peaks of the solution is given in Fig. 8.13. The energies of the vibronic peaks are plotted in Fig. 32 and a good straight line is obtained. The linear fit according to eqn 8.3 gives the vibrational energy as 0.16 eV. In the case of the crystal, we have first to identify the various peaks in the absorption spectrum. A strong vibronic progression with energies of 3.13, 3.30, 3.46 and 3.61 eV is observed in the data. These can be identified as

86



IP: 129.82.195.78

4.4 solution Energy (eV)

Name: Tapas Banerji

crystal

4.0 3.6 3.2 0

2 4 Vibronic number

Figure 32: Analysis of the vibronic peaks of anthracene shown in Fig. 8.13 as considered in Exercise 8.11. the 0–0, 0–1, 0–2, and 0–3 transitions, and a linear fit according to eqn 8.3 gives the vibrational energy as 0.18 eV. (See Fig. 32.) Two other features can also be identified in the absorption spectrum of the crystal at 3.18 and 3.35 eV. These have the same splitting as the other progression and therefore involve similar types of vibrations. The most probable cause is a splitting of the electronic states in the crystal due to its lower symmetry, eg by the Davydov effect. (See Pope and Swenberg, Electronic processes in organic crystals and polymers, 2nd edn, Oxford University Press, 1999, Section I.D.5, pp 59–66.) (8.12) When the ‘mirror symmetry’ rule works, we expect the emission spectrum to 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 band extending downwards from the 0-0 transition at 3.13 eV. The width of the band will be about 1 eV. A series of vibronic peaks will occur with energies given by hν ≈ (3.13 − n~Ω), with ~Ω ≈ 0.18 eV. We would thus expect peaks at 3.13, 2.95, 2.77, 2.59 eV · · · . (8.13) The S1 absorption band has a 0–0 transition at 1.9 eV and extends to ∼ 2.8 eV. The emission band would thus have a 0–0 transition at 1.9 eV and extend down to about 1.0 eV. The 0–1 vibronic peaks occurs at 2.1 eV in the absorption spectrum, which implies a vibrational energy of ∼ 0.2 eV, and hence a 0–1 transition in emission at around 1.7 eV. (8.14) As discussed for the PDA data in Fig. 8.16, the difference between the absorption and photoconductivity edges is caused by excitonic effects. The absorption edge corresponds to the creation of tightly-bound (Frenkel) excitons. Since excitons are neutral particles, they do not contribute to the photoconductivity. The photoconductivity edge therefore corresponds to the band edge where free electrons and holes are first created. The difference in the two edges gives the exciton binding energy, which works out to be 1.1 eV. It is important to realize that this is a different situation to that encountered for weakly-bound (Wannier) excitons. Weakly-bound excitons can 87


Name: Tapas Banerji


IP: 129.82.195.78

be easily ionized to produce free electrons and holes, and hence produce a photocurrent. (See, for example, Figs 4.5 and 6.15.) (8.15) The dominant vibrational frequency can be deduced by analysing the vibronic progression of either the absorption or emission spectra in Fig. 8.17 according to eqns 8.3 or 8.5 as appropriate. This gives ~Ω ∼ 0.17 eV, implying Ω ∼ 2.6 × 1014 rad/s. With ω = 2.98 × 1015 rad/s, we then find ωRaman = (ω − Ω) = 2.72 × 1015 rad/s, which is equivalent to a wavelength of 693 nm. Two points could be made here: • The molecule will have other vibrational modes, and these will give additional Raman lines. • Not all vibrational modes are Raman-active, (see Section 10.5.2) and it is not immediately obvious that the 0.17 eV mode responsible for the vibronic spectra will show up in the Raman spectrum. In fact, these modes are observed in the experimental Raman spectra, but it requires a careful analysis by group theory to prove that this is so. (8.16) Optical excitation creates only singlets because the ground state is a singlet and optical transitions do not change the spin. With only singlet states excited, the recombination of the electrons and holes is optically allowed for all the carriers. Wave function ↑e ↑h √ (↑e ↓h + ↓e ↑h )/√2 (↑e ↓h − ↓e ↑h )/ 2 ↓e ↓h

Sz +1 0 0 −1

S 1 1 0 1

State triplet triplet singlet triplet

Table 6: Possible arrangements of relative electron-hole spins as discussed in Exercise 8.16. On the other hand, with electrical injection there is no control of the relative spin of the electrons and holes. The spins can be either parallel or anti-parallel, and this gives rise to four possible total spin wave functions, as indicated in Table 6. Three of these are triplets and only one is a singlet state. The relative number of triplet and singlet excitons created by electrical injection is therefore in the ratio 3:1, which implies that only 25% of the excitons are in singlet states. The remaining 75% are in triplet states with very low emission probabilities. Hence the emission is expected to be weaker than that for optical excitation by a factor of four. The creation of triplets in electrically-driven organic LEDs is a serious issue that limits their efficiency. One way to enhance the efficiency is to increase the spin–orbit interaction to encourage inter-system crossing. This is typically done by including a heavy metal atom in the molecule. (See Exercise 8.9.) (8.17) (a) As we have seen in Exercise 8.16, we expect that 75% of the excitons created will be in triplet states with very low emission probabilities. Hence

88


Name: Tapas Banerji


IP: 129.82.195.78

the maximum quantum efficiency that we can expect corresponds to the number of singlet excitons that we create, namely 25%. (b) The number of electrons and holes flowing into a device carrying a current i is equal to i/e. The quantum efficiency is defined as the ratio of photons out to electrons in, and so the number of photons emitted will be equal to ηi/e. The power emitted is then equal to hν × ηi/e, and we have: P = 2.25 eV × 25% × 10 mA/e = 5.6 mW . (c) The electrical power consumed by the device is equal to iV = 50 mW. The power conversion efficiency is thus equal to 5.6/50 = 11 %. The efficiency of a real device would be much lower, mainly due to the difficulty of collecting the photons, which are emitted in all directions (ie over 4π solid angle). Only a small fraction of these would be collected by the optics. This latter point is exacerbated by the high refractive index of the molecular material, which tends to limit the effective collection efficiency even further. (See Exercise 5.13.)

a 30°

c

a

q n1 a

a1

n 2a 2

1

60°

a2 (a)

n2 |a2| sin 60°

(b)

n2 |a2| cos 60°

Figure 33: (a) Analysis of the lattice vectors of graphene, as considered in in Exercise 8.18(a). (b) Evaluation of the chiral angle θ, as required in Exercise 8.18(c). (8.18) (a) The unit cell of graphene and its lattice vectors are shown in Fig. 33(a). From inspection of this figure, we see that: √ √ |a1 | = 2 × a cos 30◦ = 2a 3/2 = 3a , where a is the length of the hexagon edge. The result for |a2 | is identical by symmetry. (b) We first find the length of the chiral vector c defined in eqn 8.14 by evaluating its scalar product: c2 ≡ c · c = (n1 a1 + n2 a2 ) · (n1 a1 + n2 a2 ) , = n21 a1 · a1 + 2n1 n2 a1 · a2 + n22 a2 · a2 . Now a1 · a1 = a2 · a2 = a20 and a1 · a2 = |a1 | |a2 | cos 60◦ = a20 /2. Hence c2 = a20 (n21 + n1 n2 + n22 ) .

89


Name: Tapas Banerji


IP: 129.82.195.78

Since c is the circumference, the tube diameter d is then given by: q c a0 d= = n21 + n1 n2 + n22 , π π as required. (c) The chiral angle for a nanotube with the chiral vector defined in eqn 8.14 can be evaluated by reference to Fig. 33(b). It is apparent that: tan θ =

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

On noting that |a1 | = |a2 |, this simplifies to: √ √ n2 sin 60◦ 3n2 /2 3n2 tan θ = = = . n1 + n2 cos 60◦ n1 + n2 /2 2n1 + n2 Equation 8.16 follows immediately. (8.19) (a) Symmetry requires that the electron wave function should be singlevalued on rotating the tube by 2π. The phase change of the electron wave function for a rotation through 2π is given by: ∆φ = k⊥ |c| ≡ k · c where k⊥ is the component of the wave vector in the direction perpendicular to the tube axis and c is the chiral vector. This must be equal to an integer multiple of 2π in order that the electron wave function is single-valued. Hence we require: k · c = 2πm , where m is an integer. (b) The tube will be metallic if the k vector corresponding to the K point of the Brillouin zone is one of the allowed wave vectors of the nanotube. This will be the case if the condition derived in part (a) is satisfied when k corresponds to the K point, that is if: K · c = 2πm , where K = (k1 − k2 )/3. Now the reciprocal lattice vectors are defined by: a1 · k1 = a2 · k2

= 2π ,

a2 · k1 = a1 · k2

= 0.

Hence K · c = (k1 − k2 ) · (n1 a1 + n2 a2 )/3 = (2πn1 − 2πn2 )/3 , and therefore 2π(n1 − n2 )/3 = 2πm , which implies n1 − n2 = 3m. 90


Name: Tapas Banerji


IP: 129.82.195.78

(8.20) We treat the nanotube as a 1–D quantum wire of length L with its axis along the z direction, and apply periodic boundary conditions so that Ψ(x, y, z) = Ψ(x, y, z + L) . The electrons are assumed to have free motion along the z axis and so their wave function is of the form: Ψ(x, y, z) = ψ(x, y) exp ikz . The periodic boundary condition requires that: exp ik(z + L) = exp ikz . This means that exp ikL = 1, and hence that k = 2πm/L, where m is an integer. The density of states in k space is therefore L/2π. We can thus write the density of states per unit length in k space as g(k) = 1/2π . The density of states per unit length in energy space is worked out from: g(E) = 2 ×

2g(k) . dE/dk

The additional factor of two here compared to eqn 3.14 comes from the fact that the +k and −k velocity states are degenerate. We assume that we have a free electron moving in a band with energy En associated with the quantized motion in the (x, y) directions. The total energy is then given by: ~2 k 2 E(k) = En + , 2m∗ so that µ 2 ¶1/2 dE ~2 k 2~ = ∗ = (E − En )1/2 . dk m m∗ The density of states is thus given by: g(E) = 4

1 2π

µ

√

¶1/2

m∗ 2~2

(E − En )−1/2 =

2m∗ (E − En )−1/2 , π~

as required for eqn 8.20. (8.21) The radiative quantum efficiency is given by eqn 5.5. The decay route via intersystem crossing to the T1 level is non-radiative, and so we can take τNR = 1.2 ns. The radiative lifetime τR is given as 1.8 µs. Hence: ηR =

1 1 = = 6.7 × 10−4 . 1 + τR /τNR 1 + (1.8 × 10−6 /1.2 × 10−9 )

The radiative quantum efficiency is therefore only 0.07%.

91


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 9 Luminescence centres (9.1) The solution for a one-dimensional potential well with infinite barriers is given in Section 6.3.2. In a cube the motion is quantized in all three dimensions, and the energies for the x, y and z directions just add together. For each direction, we have (cf eqn 6.13): E=

~2 π 2 n2 2m∗ (2a)2

where n is the quantum number. The quantum numbers for the three degrees of freedom are independent of each other, and we derive eqn 9.4 by adding the quantized energies for the x, y and z directions together. Note that the similar case of a quantum dot was considered previously in Section 6.8.1. See especially eqn 6.54. (9.2) Equation 9.5 predicts E = 0.28/a2 . The experimental energies are lower because a real F-centre is not a rigid cubic box, and hence it would be more appropriate to use a finite rather than infinite potential well model. As discussed in Section 6.3.3, the quantization energies of finite potential wells 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. Alternatively, we can just read hν ≈ 2 eV from Fig. 9.4. 2a

+

-

+

-

+

-

+

-

-

+

-

+

-

+

-

+

+

-

+

+

-

+

-

b 2a b

+ F2+ + - + - + + - + - + - + - + - + - + - +

-

2b

(a)

(b)

Figure 34: (a) An F+ 2 centre in an alkali halide with cell size 2a, as considered in Exercise 9.4. The colour centre is modelled as a rectangular box with square cross-section as shown in part (b). (9.4) The energy of the electron will be given by eqn 6.54 with dx = dy = b and

92


Name: Tapas Banerji


IP: 129.82.195.78

dz = 2b. We therefore have: ~2 π 2 E= 2m0

Ã

n2y n2x n2z + + b2 b2 4b2

! .

We can apply this model to the F+ 2 centre by taking the appropriate value of b. The rectangle equivalent to the F+ 2 centre is sketched in Fig. 34. If √the cubic unit cell size is 2a, then the square end of the box has dimension √ √ 2a, and the longer dimension is 2 2a. We therefore need to take b = 2a for an F+ → 2 transition, 2 centre. The lowest energy transition is the nz = 1 √ which has an energy of hν = 3h2 /32m0 b2 . With b = 2a, this gives hν = 3h2 /64m0 a2 , which is half the value given in eqn 9.5. The experimental absorption peak of KF:F+ 2 occurs at 1.1 eV (see Fig. 9.5), while that of KF:F occurs at 2.9 eV (see Fig. 9.4.) The experimental ratio is thus ∼ 0.4, which is close to the predicted value of 0.5. This is remarkably good agreement considering the simplicity of the model. (9.5) (a) By applying Boltzmann’s law to the two levels, we see that the occupancy of the MS = ±1 upper level relative to the MS = 0 lower level is given by: µ ¶ N2 g2 ∆E = exp , N1 g1 kB T where g2 and g1 are the level degeneracies (i.e. g2 = 2, g1 = 1), and ∆E/h is given in Fig 7.9(a) as 2.9 GHz. We thus find N2 /N1 = 1.87 for T = 2 K, which means that 1/2.87 = 35% of the centres are in the MS = 0 level, and 65% of them are in the MS = ±1 level. (b) We require N1 /(N1 + N2 ) = 80%, which implies N2 /N1 = 0.25. We thus solve: µ ¶ N2 g2 ∆E = exp = 0.25 , N1 g1 kB T for T . This gives T = 0.07 K. (9.6) The emission spectrum shows four clear peaks within the vibronic band: the zero-phonon line at 638 nm, and three phonon sidebands at 660, 685, and 708 nm. These are assigned to multiple-phonon emission processes. A fit of these energies to eqn 9.2 is shown in Fig. 35. The phonon energy is 0.064 eV. (9.7) (a) For the 3d electron we have n = 3 and l = 2, which implies hri3d = (21/2)aH /Z, while for the 4f electron we have n = 4 and l = 3 and hence hri4f = 18 aH /Z. We thus find: 7 Z4f hri3d = . hri4f 12 Z3d As a crude estimate we use average values for the atomic numbers of the relevant series: i.e. we take Z = 25 for the transition metals, and Z = 64 for the rare earths. This implies 7 64 hri3d ∼ × ∼ 1.5 . hri4f 12 25 93


Sideband energy E (eV)


IP: 129.82.195.78

E = 1.941 - 0.0636 n

1.95 1.90 1.85 1.80 1.75 0

1 2 Sideband number n

3

Figure 35: Fit to the phonon sidebands in Fig. 9.7(b) using eqn 9.2, as required for Exercise 9.6. This shows that the 3d series is expected to have a larger radius by a factor of ∼ 1.5. It is instructive to compare the radial probability densities (i.e. r2 |R(r)|2 ) derived from the radial wave functions: µ ¶2 µ ¶ r Z 7/2 4 Zr √ R3d (r) = exp − , 3/2 3aH aH 81 30 aH µ ¶3 µ ¶ Z 9/2 1 r Zr √ R4f (r) = exp − . 3/2 4aH 768 35 aH a H

Figure 36 compares r2 R(r)2 for a 3d wave function with Z = 25 with that for a 4f wave function with Z = 64. It is apparent that the 3d orbital has the larger average radius. Radial probability density (a.u.)

Name: Tapas Banerji

5

4f (Z = 64) 4 3

3d (Z = 25)

2 1 0 0.0

0.2

0.4

0.6

0.8

1.0

1.2

r / aH Figure 36: Radial probability densities for the 3d orbitals of a transition-metal ion and the 4f orbitals of a rare-earth ion, as considered in Exercise 9.7. It should be stressed that these arguments are not totally convincing because they take no account of the screening effect of the inner shells, which 94


Name: Tapas Banerji


IP: 129.82.195.78

will reduce the effective value of Z experienced by the electron. However, the more accurate calculations of the wave functions do confirm the general trends predicted by this argument. (b) The 3d transition metal ions have lost the 4s electrons, and so the 3d electrons are the outermost orbitals. By contrast, the 4f orbitals in the rare earths are inside the filled 5s and 5p orbitals. The 5s and 5p orbitals have n = 5 and l = 0 & 1 respectively. They therefore have larger radii than the 4f shell on account of the fact that hri is proportional to n2 and decreases with l. smaller distance

z

z

x z

x, y

y x, y x

(a)

(b)

(c)

(d)

Figure 37: Discussion of p orbitals as required for Exercise 9.8. (a) A pz orbital. (b) pz , px and py orbitals in an octahedral lattice, as seen in the x-z and x-y planes. (c) pz , px and py orbitals in a uniaxially-distorted lattice, as seen in the x-z and x-y planes. (d) Splitting of the p orbitals in the distorted lattice, with negative nearest neighbours. (9.8) (a) The three p orbitals are dumb-bell shaped as shown in Fig. 37(a) for the case of the pz orbital. In an octahedral lattice, the x, y and z directions are all equivalent. This implies that the px , py and pz orbitals must all experience the same interaction with the crystal. This is apparent from Fig. 37(b), which shows that the distance from the electron cloud to the ions is the same for the pz , px and py orbitals. Hence they will experience identical Coulomb interactions. (b) In a uniaxial crystal, the octahedral symmetry is lost and the z direction is now different. This means that the pz orbitals are closer to the ions than the px or py orbitals. (See Fig. 37(c).) The Coulomb interactions between the electron cloud will now be different for the pz orbital and the other two, and so its energy will be different. On the other hand, the x and y directions remain equivalent (see lower half of Fig. 37(c)), and so the px and py orbitals are still degenerate. Hence the triplet p state splits into a singlet and a doublet. (c) If the nearest neighbour ions are negative, the pz electrons will experience a stronger repulsive interaction with the lattice because of the smaller distance to the ion. Hence the pz states will have a larger energy

95


Name: Tapas Banerji


IP: 129.82.195.78

than the px and py states. This will give a splitting as shown in Fig. 37(d), with the singlet at higher energy. (9.9) (a) We write the potential in the form: V (r, θ, φ) =

6α + f (r)[f1 (θ) + f2 (θ)(e+i4φ + e−i4φ )] , a

and evaluate the matrix element from: ZZZ ∗ hm|V |m0 i = ψm V ψm0 r2 sin θ dr dθ dφ . It is immediately obvious that the constant term 6α/a can only give nonzero terms if m0 = m by wave function orthogonality. Consider the integral over φ for the remaining terms. This is of the form: Z 2π 0 I= e−imφ [f1 (θ) + f2 (θ)(e+i4φ + e−i4φ )] eim φ dφ . 0

Now it is apparent that Z 2π 00 0 e−imφ eim φ eim φ dφ = 2πδm,(m0 +m00 ) , 0

where δi,j is the Kronecker delta function. Hence the only non-zero matrix elements are those with m0 = m, m0 = m − 4 or m0 = m + 4. P2 (b) A general state of the system may be written as Ψ = i=−2 ci |ii, which may be expressed as the column vector:   c−2 c−1     Ψ=  c0  .  c1  c2 In this basis, and with the notation introduced in the Exercise, the crystalfield Hamiltonian matrix takes the form:   A 0 0 0 D 0 B 0 0 0    Hcf =  0 0 C 0 0 . 0 0 0 B 0 D 0 0 0 A The eigenenergies and states are found by diagonalizing the Hamiltonian. It is apparent that the central part of the matrix is already diagonal, and so immediately we find three eigenstates, namely: • | − 1i with energy B, • |0i with energy C, • | + 1i, also with energy B.

96


Name: Tapas Banerji


IP: 129.82.195.78

The states | − 2i and | + 2i separate into a submatrix, with: µ ¶µ ¶ µ ¶ A D c−2 c−2 =λ . D A c2 c2 The solutions are: c2 = ±c−2 , λ = A ± D. We thus find the final two eigenstates, namely: √ • (|2i + | − 2i)/ 2 with energy A + D, √ • (|2i − | − 2i)/ 2 with energy A − D. (c) The explicit calculation of the matrix elements A, B, C and D, and hence the demonstration that A + D = C and A − D = B, may be found in Henderson & Imbusch (1989), §2.6. √ If A+D = C, then the eigenstates |0i and (|2i+|−2i)/ 2 are degenerate, √ and if A − D = B, then the eigenstates | − 1i, | + 1i and (|2i − | − 2i)/ 2 are degenerate. We thus identify the two dγ states with the |0i and (|2i + √ | − 2i)/ 2 states, or a linear combination of them. Similarly, we identify √ the three d² states with the eigenstates | − 1i, | + 1i and (|2i − | − 2i)/ 2, or a linear combination of them. It is simplest to see the connection between these states and those given in the Exercise by working backwards from the Cartesian versions, and relating them to the spherical harmonic functions given in part (a) of the exercise. 2z 2 − x2 − y 2

= 2r2 cos2 θ − r2 sin2 θ cos2 φ − r2 sin2 θ sin2 φ , = r2 (2 cos2 θ − sin2 θ[cos2 φ + sin2 φ]) , = r2 (2 cos2 θ − sin2 θ) , = r2 (3 cos2 θ − 1) , ∝ C2,0 (θ, φ) , ∝ |0i ;

x2 − y 2

= = = = ∝ ∝

r2 sin2 θ cos2 φ − r2 sin2 θ sin2 φ , r2 sin2 θ(cos2 φ − sin2 φ) , r2 sin2 θ cos 2φ , r2 sin2 θ (ei2φ + e−i2φ )/2 , (C2,2 + C2,−2 ) , √ (|2i + | − 2i)/ 2 .

We thus identify the wave functions 2z 2 − x2 − y 2 and x2 − y 2 with the two dγ states. Similarly: xy

= r2 sin2 θ cos φ sin φ , = r2 sin2 θ sin 2φ/2 , = r2 sin2 θ (ei2φ − e−i2φ )/4i , ∝ (C2,2 − C2,−2 ) , √ ∝ (|2i − | − 2i)/ 2 ; 97


Name: Tapas Banerji


IP: 129.82.195.78

yz

= r2 sin θ sin φ cos θ , = r2 cos θ sin θ(eiφ − e−iφ )/2i , ∝ (C2,1 − C2,−1 ) , √ ∝ (|1i − | − 1i)/ 2 ;

zx

= r2 cos θ sin θ cos φ , = r2 cos θ sin θ(eiφ + e−iφ )/2 , ∝ (C2,1 + C2,−1 ) , √ ∝ (|1i + | − 1i)/ 2 .

We thus identify the wave functions xy, yz and zx with the three d² states.

y

y

(a)

z

(b)

(c)

x

x

x

q = -Ze d xy

d x2 - y 2

d z2

Figure 38: Probability densities for the d² and dγ states, as considered in Exercise 9.9(d): (a) dxy , (b) dx2 −y2 , and (c) dz2 . The circles on the axes represent the anions, which have charge −Ze. The probability densities for the dyz and dzx states are obtained by exchanging the axis labels. (d) The probability densities for the dγ and d² states are plotted in Fig. 38. The dγ state with wave function 2z 2 − x2 − y 2 is labelled dz2 . The dxy , dyz and dzx states have zero probability along the axes, and have high probability between the axes. On the other hand, the dx2 −y2 and dz2 states have high probability densities along the axes, but not along the diagonal directions. For a d1 configuration the ion has a single d electron. In the dγ states the electron probability density is high along the crystalline axes, and so the electron experiences a strong repulsion from the negatively-charged anions. The d² states, by contrast, have a smaller probability density close to the anions and thus experience a smaller repulsion. The two dγ states therefore have the higher energy, and hence the doublet lies above the triplet. The argument is reversed for a d9 configuration, which can be considered as a single hole in the filled d shell. The charge cloud is now positive rather than negative, and the two dγ states experience a stronger attraction rather than repulsion. They therefore have the lower energy.

98


Name: Tapas Banerji


IP: 129.82.195.78

(9.10) The 1064 nm line in a Nd:YAG crystal corresponds to a transition from the 11 502 cm−1 level as shown in Fig. 9.9(b). The relative populations of the 11 502 cm−1 and 11 414 cm−1 levels of the 4 F3/2 term are proportional to the Boltzmann factor: µ ¶ N (11502 cm−1 ) ∆E = exp − , N (11414 cm−1 ) kB T where ∆E = 88 cm−1 . This ratio equals 0.19 at 77 K and 0.66 at 300 K. The spontaneous emission rate increases in proportion to these factors, and therefore the relative intensity of the 1064 nm line increases with T . (9.11) The transition rates between the two levels are governed by the Einstein coefficients, 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. For gain, we need that the stimulated emission rate should exceed the absorption rate. (Spontaneous emission is negligible at high light intensities.) The condition for this to occur is (see eqns B.5 and B.6): B21 N2 u(ν) > B12 N1 u(ν) , which implies:

N2 B12 > . N1 B21

On substituting from eqn B.10, we derive the condition for net gain: N2 g2 > , N1 g1 where g2 and g1 are the degeneracies of the two levels. This condition is called population inversion. (9.12) (a) When the pump is turned off, all of the atoms will be in the ground state, so that: N0 N2

= N, = 0,

where N is the total number of atoms. When the pump is turned on, ∆N atoms will be pumped to the upper laser level via level 1, so that: N0

=

N − ∆N .

N2

= ∆N .

For population inversion, we require N2 > N0 , which implies ∆N > (N − ∆N ), i.e. ∆N > N/2. Population inversion is therefore only achieved when more than 50% of the atoms are pumped to the excited state. (b) In this particular example, we have N2 = 0.6N at t = 0, and we have seen in part (a) that the laser will stop oscillating when N2 = 0.5N . The sequence of events is therefore as shown in Fig. 39. The number of atoms that make stimulated radiative transitions (and hence the number 99


Name: Tapas Banerji


IP: 129.82.195.78

of photons emitted) during the laser pulse is therefore 0.1N . This gives a pulse energy of 0.1N × hν, where hν is the laser photon energy, namely 1.79 eV. On inserting the appropriate numbers we find: Epulse = 0.1 × (1025 × 10−6 ) × 1.79 eV = 0.3 J .

60%

50%

2

2 laser pulse emitted 40%

0

50% 0

t=0

after pulse

Figure 39: Sequence of events for the ruby laser considered in Exercise 9.12(b). (9.13) (a) The optical intensity I(t) is proportional to |E(t)|2 , and hence if we have a Gaussian pulse as defined in the exercise, we have a time-varying electric field of the form (note the extra factor of two): E(t) = E 0 exp(−t2 /2τ 2 ) e−iω0 t , where ω0 is the centre angular frequency. The spectrum of the pulse is found by first taking the Fourier transform of E(t): Z +∞ 1 E(ω) = √ E(t) eiωt dt , 2π −∞ Z +∞ E0 √ exp(−t2 /2τ 2 ) ei(ω−ω0 )t dt , = 2π −∞ = E 0 τ exp[−τ 2 (ω − ω0 )2 /2] , where we used the standard integral: Z +∞ √ F (Ω) = exp(−t2 /σ 2 )e−iΩt dt = πσ exp(−σ 2 Ω2 /4) , −∞

in the last line. The final result is obtained by using: I(ω) ∝ |E(ω)|2 ∝ exp[−τ 2 (ω − ω0 )2 ] . (b) The full width at half maximum (FWHM) of the pulse in the time domain is found by finding the times for which I(t) = I0 /2, i.e. by solving: exp(−t2 /τ 2 ) = 0.5 . √ √ This gives t = ± ln 2 τ , so that ∆t = 2 ln 2 τ . The FWHM of the pulse in the frequency domain is likewise found by solving: exp[−τ 2 (ω − ω0 )2 ] = 0.5 , 100


Name: Tapas Banerji


IP: 129.82.195.78

√ √ which gives (ω − ω0 ) = ± ln 2/τ , and hence ∆ω = 2 ln 2/τ . We then obtain: ∆ν∆t = ∆ω∆t/2π = 2 ln 2/π = 0.441 . (9.14) The crystal-field shifts of the energy levels depend on the local environment of the ion. For example, small perturbations to the atomic positions affect the energy levels of the ion through the alterations to the local electric field the ion experiences. There will be much larger inhomogeneity in the local environment in a glass than in a crystal, due to the lack of longrange order. Hence we expect much larger inhomogeneous broadening of the crystal-field split transitions in a glass than in a crystal. If we assume a Gaussian pulse, we expect a time-bandwidth product of 0.441, and hence ∆tmin = 60 fs. Other pulse shapes would give comparable minimum pulse durations. (9.15) The subscript ‘g’ stands for gerade and implies even parity. A g→g transition therefore involves no parity change and is forbidden for electricdipole transitions. (See Section B.3.) The transition will therefore have a lower probability than for an electric-dipole allowed process, and hence have a long excited state lifetime (∼ 4 µs). Since the lifetime is long, the radiation would be classified as phosphorescence rather than fluorescence. (See Section B.3.) (9.16) The excited state lifetime is determined by both radiative and nonradiative processes. It follows from eqn 5.4 that: 1 1 1 = + . τ τR τNR The radiative lifetime is governed by atomic transition probabilities and is not expected to vary significantly with the temperature. On the other hand, the non-radiative transition rate is governed by phonon-assisted processes and is expected to increase strongly with T . On substituting τR = 1.8 ms into the equation above, we find τNR = 6.3 ms at 77 K and 0.062 ms at 300 K. This implies, through eqn 5.5, that the radiative quantum efficiency is 78% at 77 K and 3% at 300 K. The radiative efficiency is too low at 300 K to allow lasing. (9.17) The level scheme for Ti:sapphire is shown in Fig. 9.13. If we assume that the laser threshold is low, and that the slope efficiency is 100% (i.e. one laser photon emitted for each pump photon absorbed), then the ratio of the output power to the input power would just be proportional to the ratio of the respective photon energies, which implies: Pout =

1/800 hνout Pin = Pin = 3.2 W . hνin 1/514

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 in the crystal even for the ideal case of 100% quantum efficiency due to the difference in the photon energies of the argon and Ti:sapphire lasers.

101


Name: Tapas Banerji


IP: 129.82.195.78

In reality, a Ti:sapphire laser typically gives about 1 W for a pump power of 5 W at ∼ 500 nm. This reduction of the power from the ideal value is caused by a number of factors: • The laser threshold will be significant, and the output power is proportional to (Pin − Pth ) rather than to just Pin (cf. Fig 5.15(a)); • the quantum efficiency is less than 100% due to significant nonradiative decay;1 • the absorption of the pump laser in the crystal is not perfect, and so not all of the pump power is absorbed; • there might be optical losses within the cavity. (9.18) The energy conversion efficiency is calculated by assuming that we obtain one emitted photon from the phosphor for each photon absorbed from the LED. The intrinsic energy conversion efficiency is then just determined by the ratio of the photon energies: η=

hνout λin = , hνin λout

where λin and λout are the wavelengths of the photons absorbed and emitted, respectively. In this case we have λout = 650 nm. Hence: (a) η = 54% for λin = 350 nm; (b) η = 69% for λin = 450 nm.

1 The

operating temperature of the laser crystal will be above room temperature due to the heat generated within it, and this further increases the non-radiative decay rate. Cooling of the laser rod is therefore essential.

102


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 10 Phonons (10.1) The phonon modes of purely covalent crystals do not give rise to infrared absorption because the atoms are neutral and do not interact with the electric field of the light wave. Of the five materials listed, germanium and argon are elemental, and must therefore have neutral atoms with nonpolar bonds, and hence no infrared absorption. The other three, namely ice, ZnSe and SiC, are polar, and would therefore have some infrared-active phonons. (10.2) It is apparent from eqn 1.29 that R = 0 when n ˜ = 1, and hence ²r = 1. For an undamped oscillator, ²r (ν) is given by eqn 10.15. We thus solve: ²r (ν) = ²∞ + (²st − ²∞ )

2 νTO 2 − ν2) = 1 , (νTO

for ν. Rearrangement gives: µ 2

ν =

²st − 1 ²∞ − 1

¶ 2 νTO ,

leading to the result quoted in the exercise. (10.3) The exercise follows Example 10.1(a). We calculate νLO = 20 THz from the LST relationship, and hence that the Reststrahl band runs from 9.2 to 20 THz, i.e. 15 µm to 33 µm. (10.4) The exercise closely follows Example 10.1(b). The relative permittivity is given by eqn 10.10 as: ²r (ν) = 10 +

210 , 100 − − iγ 0 ν/2π ν2

where ν is measured in THz and γ 0 = γ/1012 . We calculate νLO = 11 THz from the LST relationship, so that the Reststrahl band runs from 10 to 11 THz. We thus need to evaluate ²r at ν = 10.5 THz. (a) With γ = 1011 s−1 , we find ²r = −10.48+0.336i at 10.5 THz, and hence that n = 0.0519 and κ = 3.238 from eqns 1.25–26. Then from eqn 1.29 we find R = 0.98. (b) With γ = 1012 s−1 , we find ²r = −9.958+0.252i and n ˜ = 0.509+3.196i at 10.5 THz, and hence that R = 0.84. (10.5) (a) We identify the Reststrahl band from the region of high reflectivity from 30–32 µm. (See Fig. 40(a).) On equating the upper and lower wavelength limits with νTO and νLO respectively, we find νTO ≈ 9.5 THz and νLO ≈ 10 THz. 103



nLO

IP: 129.82.195.78

nTO 1.00

0.8

Reflectivity

1.0 Reflectivity

Name: Tapas Banerji

(a) AlSb

0.6 0.4

R(¥)

R(0) 0.2 0.0 16

(b)

0.96 0.92 0.88 0.84

20

24 28 32 36 Wavelength ( mm)

40

0.0

0.4

0.8 1.2 g (1012 s-1)

1.6

Figure 40: (a) Interpretation of the data in Fig. 10.14 as required for Exercise 10.5(a) and (b). (b) Calculated reflectivity versus damping constant, as required for Exercise 10.5(c). (b) The high- and low-frequency permittivities can be deduced from the asymptotic reflectivities. (See Fig. 40(a).) The low-frequency limit gives √ ²st from (see eqn 1.29, with n = ²r ): R(0) =

µ√ ¶ ²st − 1 2 . √ ²st + 1

At ω → 0, there is no absorption, and so ²r will be real. On reading R ≈ 0.30 ≡ R(0) at long wavelengths, we deduce ²st ≈ 12. On similarly equating the short wavelength limit of R, namely 26%, with R(∞), we deduce ²∞ ≈ 9.5. (c) The peak reflectivity is about 90%, and is limited by γ, which in turn is determined by the lifetime of the TO phonons. The middle of the Reststrahl band occurs at 9.75 THz. We thus need to evaluate ²r from (see eqn 10.10, with ν measured in THz and γ 0 = γ/1012 ): ²r (ν) = ²∞ + (²st − ²∞ )

2 (νTO

2 νTO 225 = 9.5 + , 2 2 − ν ) − iγν/2π 90 − ν − iγ 0 ν/2π

at ν = 9.75 THz. We split this into the real and imaginary parts, compute n and κ from eqns 1.25–26, and R from eqn 1.29. The reflectivity calculated in this way is plotted as a function of γ in Fig. 40(b). It is apparent that we have R = 0.9 for γ = 8.6 × 1011 s−1 . This implies, from γ = 1/τ , that τ = 12 ps. The values of νLO and νTO found in part (a) can be compared to the Lyddane–Sachs–Teller relationship, which predicts νLO /νTO = 1.11. The experimental ratio is slightly smaller. The values given here are only approximate, and depend on how exactly they are extracted from the data. The departure from LST is therefore not very significant. (10.6) The exercise closely follows Example 10.1(c). We first use eqn 10.17 to find ²r at νTO , which gives ²r = 10 + 132i for γ = 1012 s−1 and ²r = 10 + 1320i for γ = 1011 s−1 . We then use eqn 1.26 to find κ and 104


Name: Tapas Banerji


IP: 129.82.195.78

eqn 1.19 to find α. (a) For γ = 1012 s−1 we find κ = 7.8 and α = 3.4 × 106 m−1 . (b) For γ = 1011 s−1 we find κ = 26 and α = 1.1 × 107 m−1 . Note that the peak absorption increases for the smaller value of the damping, as normal for a damped oscillator. (10.7) The peak reflectivity is governed by the damping constant γ. (See, for example, Fig. 40(b) above.) As the temperature increases, we expect γ to increase, and hence R to decrease, due to the increased probability of anharmonic decay processes of the type illustrated in Fig. 10.13. The reason why anharmonic phonon decay increases with T is that phonons are bosons, and the probability for phonon emission increases as the thermal population of the final state increases. (n.b. This contrasts with fermions, for which the transition probabilities decrease with increasing occupancy of the final state.) (10.8) With negligible damping, we can use eqn 10.15 to calculate ²r = 21.5 at 8 THz. We then substitute this value of ²r into eqn 10.18 to compute the wave vector. This gives: √ √ ²r ω 21.5 × 2π × 8 × 1012 q= = = 7.8 × 105 m−1 . c 3 × 108 (10.9) (a) The condition for cyclotron resonance is given in eqn 10.24. In a polar material, the mass that is measured is the polaron mass m∗∗ . We thus obtain: eλB m∗∗ = = 0.097 m0 . 2πc (b) The rigid lattice mass m∗ can be calculated from eqn 10.21. On inserting the relevant values into eqn 10.19, we find αep ≈ 0.33 for CdTe. Then from eqn 10.21 we find that m∗∗ = 0.097 m0 implies m∗ = 0.092 m0 . (10.10) The Raman spectra for a number of III-V crystals are shown in Fig. 10.11. In each case we observe two peaks: one for the TO phonons and the other for the LO phonons. These two phonon modes have different frequencies because III-V compounds have polar bonds with partially charged atoms. They therefore interact with light, and obey the LST relationship. The situation in diamond is different because it is a purely covalent crystal, with neutral atoms that do not interact with the light. The LST analysis does not apply, and the optical phonons are degenerate at q = 0. The Raman spectrum therefore has only one peak for the optical phonons. (10.11) Silicon, like diamond in the previous exercise, is covalent, and its LO and TO phonons are degenerate at q = 0. The two peaks correspond to the 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 and is thus the anti-Stokes line, while that at 528.6 nm is the Stokes line. The phonon frequency can be worked out from eqn 10.27, which gives, for the case of the Stokes line: Ω/2π =

c c − = 15.5 THz . 514.5 nm 528.6 nm 105


Name: Tapas Banerji


IP: 129.82.195.78

The relative intensities of the lines are given by eqn 10.29: µ ¶ I(501.2 nm) h × 15.5 × 1012 = exp = 0.08 . I(528.6 nm) kB T (10.12) NaCl has an inversion centre and so the mutual exclusion rule applies. It is apparent from Fig. 41 that the TO mode has odd parity under inversion. The TO mode is therefore IR active but not Raman active.

invert

Figure 41: Inversion of a TO mode in an ionic crystal, as considered in Exercise 10.12. The inversion centre is circled. (10.13) The energies of the phonon modes can be deduced directly from the Raman spectra by applying eqn 10.27, with the + sign as appropriate for a Stokes shift. This shows that the Raman shift is exactly equal to the phonon frequency. For each crystal, two lines are observed. The lower frequency line comes from the TO phonons, while the higher frequency line originates from the LO phonons. The Raman shifts in cm−1 from Fig. 10.11 are given in Table 7, together with the energies deduced according to: E (meV) = 0.124 × Raman shift (cm−1 ) . Crystal GaAs InP AlSb GaP

Raman line 1 cm−1 262 299 312 364

Raman line 2 cm−1 286 341 332 403

TO phonon energy meV 32.5 37.1 38.7 45.1

LO phonon energy meV 35.5 42.3 41.1 50.0

Table 7: Raman shifts deduced from the data in Fig. 10.11, as considered in Exercise 10.13. On comparing the frequencies of the TO and LO phonons of GaAs in Table 7 with those deduced from the infrared reflectivity data in Fig. 10.5, we see that there is a small shift of a few wave numbers between the two sets of data. This is caused by the slight decrease of the optical phonon frequencies between 4 K and 300 K. (10.14) Equation 10.28 implies that momentum is conserved during the Raman scattering process so that the vectors form a triangle as depicted in Fig. 42(a). In the case of inelastic scattering by acoustic phonons, the frequency shift of the photon is very small because ω À Ω. This implies that the magnitude of the photon wave vector hardly changes, so that we can approximate: nω . |k1 | = |k2 | ≡ k = c 106


Name: Tapas Banerji


IP: 129.82.195.78

k2

q

q

k1

k2

k1

q/2 (a)

(b)

Figure 42: (a) Conservation of momentum during Raman scattering by a phonon of wave vector q, as as considered in Exercise 10.14. (b) Back-scattering geometry. It is then apparent from Fig. 42(a) that: q θ nω θ = k sin = sin . 2 2 c 2 On writing q = Ω/vs as appropriate for acoustic phonons, we derive the result in the exercise. Equation 10.32 then follows by writing: δω = |ω2 − ω1 | = Ω . (10.15) In back-scattering geometry, we have θ = 180◦ , so that q = 2k. (See Fig. 42(b) with |k1 | = |k2 | = k.) It then follows from eqn 10.32 with sin(θ/2) = 1 that: cδω λδν vs = = . 2nω 2n On inserting the data given in the exercise, we find vs = 813 m s−1 . (10.16) (a) The negative term in r−1 is the attractive potential due to the Coulomb interaction between the ions. The Madelung constant α accounts for the summation of the contributions of the positive and negative ions over the whole crystal. The positive term in r−12 represents the short range repulsive force due to the Pauli exclusion principle when the electron wave functions overlap. (b) The graph of U (r) is qualitatively similar to that for the Lennard–Jones potential, being attractive for large r, repulsive for small r, and with a minimum at some intermediate value of r, labelled r0 . (cf. Fig. 28.) The value of r0 is found by differentiating U (r): dU 12β αe2 = − 13 + , dr r 4π²0 r2 and finding the value of r for which dU/dr = 0, namely: 12β αe2 = , r013 4π²0 r02 which implies r011 = 12β × 4π²0 /αe2 . (c) The Taylor series for U (r) expanded about r0 is: U (r) = U (r0 )+

3 dU 1 d2 U 2 1d U (r−r0 )+ (r−r ) + (r−r0 )3 +· · · . 0 dr r=r0 2 dr2 r=r0 6 dr3 r=r0

107


Name: Tapas Banerji


IP: 129.82.195.78

Now U (r) has a minimum at r0 , and the first derivative is therefore zero, so that: U (r) = U (r0 ) +

1 d2 U 1 d3 U 2 (r − r ) + (r − r0 )3 + · · · . 0 2 dr2 r=r0 6 dr3 r=r0

We can reconcile this with eqn 10.33 by taking x = r − r0 and U (x) = U (r) − U (r0 ). It is then apparent that: C2

=

C3

=

1 d2 U , 2 dr2 r=r0 1 d3 U . 6 dr3 r=r0

At r = r0 we have d3 U dr3

6αe2 2184β = − 15 + , r 4π²0 r04 µ 02 ¶ 6αe 2184β 1 , = − 11 4π²0 r0 r04 176αe2 = − , 4π²0 r04

where we used the result of part (b) to derive the last line. Hence: C3 = −22αe2 /3π²0 r04 . (10.17) If the Raman spectrum is lifetime-broadened, we shall have a Lorentzian line shape with: 1 ∆ν ∆t = . 2π 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 value agrees with the lifetime measured by time-resolved Raman scattering. See: von der Linde et al., Phys. Rev. Lett. 44, 1505 (1980).

108


Name: Tapas Banerji


IP: 129.82.195.78

Chapter 11 Nonlinear optics (11.1) In the Bohr model for hydrogen, the radius of the electron in the nth quantum level is given by: rn =

4π²0 ~2 n2 n2 = aH . me2 Z Z

The magnitude of the electric field is given by the standard Coulomb formula: Ze E= , 4π²0 r2 which, on inserting rn from the Bohr formula, gives: E=

e Ze Z3 = . 4π²0 rn2 4π²0 a2H n4

For the outer 3s and 3p electrons in atomic silicon, use n = 3 and an effective nuclear charge Z ∼ 4.2 We then obtain a value of E = 5 × 1011 V m−1 . The field for the conduction electrons in crystalline silicon would, of course, be different due to the high relative permittivity and the low effective mass. (11.2) We can relate the optical intensity to the electric field by using eqn A.44. (a) The optical intensity is found from: I=

P Epulse /τpulse 1 ÷ 10−8 2 = = = 5.1 × 1012 W/m . A πr2 π(2.5 × 10−3 )2

Hence with n = 1 for air, we find from eqn A.44 that E = 6.2×107 V m−1 . (b) The optical intensity is found from: I=

P 10−3 2 = = 5 × 107 W/m . A 20 × 10−12

Then with n = 1.45 as appropriate for the fibre, we find from eqn A.44 that E = 1.6 × 105 V m−1 . (11.3) With no external field applied, the gas is isotropic and therefore possesses inversion symmetry. Hence χ(2) = 0, and no frequency doubling will occur. With the electric field applied, the gas is no longer isotropic as the electron clouds 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 and frequency doubling can, in principle, occur. However, it would give a very weak signal due to the low density of atoms. 2Z eff

is the difference between the nuclear charge and the total number of inner shell electrons that screen the valence electrons from the nucleus. i.e. Zeff = 14 − 10, where 10 is the total number of electrons in the 1s, 2s and 2p shells.

109


Name: Tapas Banerji


IP: 129.82.195.78

(11.4) The second-order nonlinear susceptibility is zero if the material has an inversion centre. We must therefore consider the microscopic structure to see 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 diamond structure except that the bonds are asymmetric. It does not possess inversion symmetry and therefore χ(2) 6= 0. (c) Water is a liquid and is therefore isotropic; hence inversion symmetry applies and χ(2) = 0. (d) Glass is amorphous and has no preferred axes; inversion symmetry applies 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 inversion symmetry. Hence χ(2) 6= 0. (11.5) (a) Consider the absorption and stimulated emission transitions as indicated in Fig. 11.2. (Spontaneous emission can be neglected if uν is sufficiently large.) The absorption and stimulated emission rates are equal to N1 B12 uν g(ν) and N2 B21 uν g(ν) respectively. (See eqns B.5–6 with the additional factor of g(ν) explained in the discussion of eqn 11.30.) If the levels are non-degenerate, then eqn B.10 tells us that B12 = B21 . At t = 0, all the atoms are in level 1, and there is net absorption, which increases N2 and decreases N1 . As the atoms are pumped to level 2, the stimulated emission rate becomes increasingly significant. Eventually, we reach a stage where N1 = N2 = N0 /2, and the stimulated emission and absorption rates are identical. There is therefore no net absorption or emission, and N2 cannot increase further. The maximum value of N2 that can be achieved is therefore N0 /2.3 (b) If we only have two levels and we neglect spontaneous emission, then the rate equations for N1 and N2 are: dN1 dt dN2 dt

= −B12 N1 uν g(ν) + B21 N2 uν g(ν) , = +B12 N1 uν g(ν) − B21 N2 uν g(ν) .

On setting B12 = B21 as appropriate for non-degenerate levels, and subtracting, we find: d∆N d (N1 − N2 ) = = −2B12 uν g(ν)∆N , dt dt where ∆N = N1 − N2 . Integration yields: ∆N (t) = ∆N (0) exp(−2B12 uν g(ν)t) , which, with ∆N (0) = N0 , gives the required result. 3 This shows that it is not possible to achieve population inversion (i.e. N > N ) in a 2 1 two-level system: three or more levels are required. This is why lasers, in which population inversion is essential, are always classified as either three or four-level systems.

110


Name: Tapas Banerji


IP: 129.82.195.78

The result quantifies the way the laser pumps atoms from level 1 to level 2, and hence reduces the net absorption. The equation implies that the populations will eventually equalize no matter how weak the laser beam is. This unphysical result arises from neglecting spontaneous emission and transitions to other levels that occur in real atoms.

y direction of propagation q

x

E

Figure 43: Propagation and polarization vectors for the light wave considered in Exercise 11.6. (11.6) With the beam propagating in the x-y plane and with its polarization in the same plane, the light must be linearly polarized as shown in Fig. 43. The z-component of the electric field is therefore zero. The nonlinear polarization, for the given nonlinear optical tensor, is found from eqn 11.43 to be:   E xE x  (2)     EyEy     Px 0 0 0 d14 0 0  2d14 E y E z  EzEz   (2)      Py  = 0 0 0 0 d25 0   2E y E z  = 2d25 E z E x .   (2) 0 0 0 0 0 d36  2d14 E y E z Pz 2E z E x  2E x E y On setting E z = 0, we obtain:  (2)    Px 0  (2)   . 0 Py  = (2) 2d E E 14 y z Pz (2)

Only Pz 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 respect to the x axis as shown in Fig. 43 and that the light has an electric field of magnitude E 0 . It will then be the case that E x = E 0 cos θ and E y = E 0 sin θ, and hence that: Pz(2) = 2d36 E 20 cos θ sin θ = d36 E 20 sin 2θ . This is maximized when 2θ = 90◦ : i.e. θ = 45◦ . (11.7) This exercise closely follows Example 11.2, and the phase-matching angle is found by substituting the appropriate refractive indices into eqn 11.52. 111


Name: Tapas Banerji


IP: 129.82.195.78

With the data given in the exercise, this gives: 1 sin2 θ cos2 θ = + . 2 2 (1.506) (1.490) (1.534)2 On using cos2 θ = 1 − sin2 θ and re-arranging, we find sin2 θ = 0.626 and hence θ = 52.3◦ . (11.8) (a) In the absence of the field, the index ellipsoid given in eqn 11.54 is of the form: x2 + y 2 z2 + = 1. n2o n2e On substituting into eqn 11.60 with E x = E y = 0 and E z = E, we find that the only non-zero changes to the index ellipsoid induced by the field are as follows: µ ¶ µ ¶ 1 1 ∆ = ∆ = r13 E , n2 1 n2 2 µ ¶ 1 ∆ = r33 E . n2 3 From eqn 11.56 we then see that the modified index ellipsoid is: ¶ µ ¶ µ ¶ µ 1 1 1 2 2 + r13 E x + + r13 E y + + r33 E z 2 = 1 , n2o n2o n2e which can be written in the form : x2 + y 2 z2 + = 1, no (E)2 ne (E)2 where: 1 n2o (E) 1 n2e (E)

= =

1 + r13 E , n2o 1 + r33 E . n2e

This implies: no (E) = ne (E) =

no (1 + n2o r13 E)−1/2 , ne (1 + n2e r33 E)−1/2 .

We assume that the field-induced changes are small and use the approximation (1 + x)−1/2 = (1 − x/2) for small x to obtain the final result: 1 no (E) = no − n3o r13 E , 2 1 ne (E) = ne − n3e r33 E . 2 (b) The geometry of the crystal is shown in Fig. 44. If the light is propagating along the y axis, and is polarized along the z direction, then it 112


Name: Tapas Banerji


IP: 129.82.195.78

L z y V

x

Figure 44: Experimental geometry of the electro–optic phase modulator considered in Exercise 11.8. will experience a refractive index of ne (E). From part (a) we know that ne (E) = ne − n3e r33 E/2, and hence the refractive index change when the field is applied. The phase change associated with the change in ne is therefore given by ∆φ(E) =

2π ∆ne L = −πn3e r33 EL/λ . λ

The argument would work equally well if the light were polarized along x instead of z. In this case the phase change would be given by: ∆φ(E) =

2π ∆no L = −πn3e r13 EL/λ . λ

In the case of LiNbO3 , r33 ∼ 3r13 (see Table 11.3), and so it is convenient to use the z polarization, but this is not necessarily true for other crystals. (c) It is apparent from part (b) that the electric field produces a linear modulation of the phase. Hence the application of a voltage modulates the phase in proportion to the voltage. This is a phase modulator.

L x x¢ y¢

z y V

Figure 45: Experimental geometry of the electro-optic crystal considered in Exercise 11.9. (11.9) We consider an electro-optic crystal with axes as defined in Fig. 45. The voltage is applied so as to produce an electric field of magnitude E z along z axis. (a) If the light propagates along the z axis, the polarization vector will lie in the x-y plane, or equivalently, in the x0 -y 0 plane. We resolve the 113


Name: Tapas Banerji


IP: 129.82.195.78

light polarization vector into its components along the x0 and y 0 axes. The phase shift induced by a refractive index change ∆n in a medium of length L is, in general, given by: 2π ∆nL , λ

∆φ =

where λ is the vacuum wavelength. On applying this to the two components along the x0 and y 0 axes, we then have: ∆Φx0

=

∆Φy0

=

2πL πLn30 r41 ∆nx0 = Ez λ λ 2πL πLn30 r41 ∆ny0 = − Ez , λ λ

where ∆nx0 and ∆ny0 are the field-induced refractive index changes along the x0 and y 0 axes as given in the exercise. The phase difference ∆Φ is thus given by (with E z = V /L): ∆Φ = ∆Φx0 − ∆Φy0 =

2πLn30 r41 E z 2πn30 r41 V = . λ λ

Note that the phase change is independent of the length in this longitudinal geometry. (b) On setting ∆Φ = π, we find: Vπ =

λ . 2n30 r41

With the appropriate figures for CdTe given in the exercise, we obtain Vπ = 44 kV. (11.10) The phase change is given by eqn 11.62. For a crystal of length L, the field strength at voltage V is equal to V /L. Hence the phase change is: ∆φ =

2π 3 n r63 (V /L)L = 2πn3o r63 V /λ . λ o

The transmission will be equal to 50% when ∆φ = π/2, since this means that the crystal acts like a quarter wave plate. In these circumstances, the output of the crystal is circularly polarized, so that half of the intensity is transmitted through the second polarizer. This occurs when: V = λ∆φ/2πn3o r63 = λ/4n3o r63 . On inserting the values of no and r63 for λ = 633 nm, we find V = 4.2 kV. (11.11) In a third-order nonlinear medium, the change in the relative permittivity is given by (see eqn 11.69): ∆˜ ²r = ∆²1 + i∆²2 = χ(3) E 2 . We therefore have ∆²2 = Im(χ(3) )E 2 and hence that ∆²2 ∝ Im(χ(3) )I because I ∝ E 2 . It follows from eqns 1.19 and 1.24 that ∆α ∝ ∆²2 . Hence, in a medium with Im(χ(3) ) 6= 0, we expect ∆α ∝ Im(χ(3) )I . 114


Name: Tapas Banerji


IP: 129.82.195.78

A saturable absorber has an absorption coefficient that obeys eqn 11.37. In the limit of small intensities, this is of the form: α(I) = α0 − α0 I/Is = α0 − ∆α , where ∆α = α2 I and α2 = α0 /Is . We thus have an intensity dependence exactly as described above, and we therefore conclude that the saturable absorber must have Im(χ(3) ) 6= 0. (11.12) We can choose our axes as we please in an isotropic medium. Therefore choose z as the direction of propagation, and x as the polarization vector, so that the electric field is given by: E = (E x , 0, 0) . The third-order nonlinear polarization is given by eqn 11.11, and, with E y = E z = 0, the nonlinear polarization is of the form:  (3)    Px χxxxx  (3)  3  Py  = ²0 E x  χyxxx  . (3) χzxxx Pz However, we see from Table 11.6 that χyxxx = χzxxx = 0. Hence we find P = ²0 χxxxx E 3x (1, 0, 0) , which means that P is parallel to E. (11.13) This exercise is very similar to Example 11.4. From eqn 11.76, we require that: 2π ∆Φnonlinear = n2 IL = π , λ which implies: I=

λ 1.55 × 10−6 = = 3.9 × 1012 W m−2 . 2n2 L 2 · 2 × 10−20 · 10

The optical power to produce this intensity is given by: P = IA = 3.9 × 1012 × π(2.5 × 10−6 )2 = 76 W . This is a large power level for a continuous-wave laser, but not for a pulsed laser. Consider, for example, a mode-locked laser with an average power of Pav , pulse repetition rate f , and pulse width tp . The peak power is given by: Pav Epulse = . Ppk = tp f tp On inserting typical values, namely Pav = 10 mW, f = 100 MHz, and tp = 1 ps, we find Ppk = 100 W.

115


Name: Tapas Banerji


IP: 129.82.195.78

(11.14) It is apparent from Fig. 11.10 that the presence of electrons causes the states in the conduction band up to EFc to be filled up, and likewise for the holes in the valence band. The absorption between Eg and (Eg + EFc + EFv ) will therefore be blocked, and the new absorption edge will occur at (Eg + EFc + EFv ). The shift in the absorption edge is therefore (EFc + EFv ). The Fermi energy in the conduction band can be calculated from eqn 5.13. On inserting m∗e = 0.067m0 , we find EFc = 0.054 eV. In the case of the valence band, we must consider the occupancy of both the heavy and light hole bands. On using the result of Exercise 5.14(b), we have: 1 Nh = 3π 2

µ

2 ~2

¶3/2

3/2

3/2

(mhh + mlh ) (EFv )3/2 ,

which gives EFv = 0.007 eV for m∗hh = 0.5m0 and m∗lh = 0.08m0 . Hence the absorption edge will shift to higher energy by 0.054 + 0.007 = 0.061 eV.4 (11.15) If we treat the exciton as a classical oscillator, we can use the results derived in Chapter 2. If we assume that the contribution of the exciton to the refractive index is small compared to the non-resonant value, (which is indeed the case, as we shall show below,) then we expect a refractive index variation as in Fig. 2.5. The refractive index will thus have local maxima and minima just below and above the centre of the absorption line. It is this extra contribution that we are considering in this exercise. The magnitude of the excitonic contribution can be calculated by following the method of Example 2.1. In part (a) of Example 2.1 it is shown that κmax =

N e2 , 2n²0 m0 γω0

where κmax is the extinction coefficient at the line centre, while in part (c) it is shown that: µ ¶1/2 N e2 nmax = ²∞ + . 2²0 m0 γω0 It then follows, with ²∞ = n2 , that: ³ κmax ´1/2 nmax = n 1 + , n where nmax is the maximum value of the refractive index. If we assume, as is demonstrated below, that κmax ¿ n, we then find: nmax = n + κmax /2 . Now we know from eqn 1.19 and the data given in the exercise that κmax =

λαmax 847 × 10−9 · 8 × 105 = = 0.054 . 4π 4π

4 In

a real experiment, the behaviour would be more complicated due to many-body effects such as band gap renormalization. This causes a shift of Eg ∝ −N 1/3 , which reduces the blue shift of the absorption edge.

116


Name: Tapas Banerji


IP: 129.82.195.78

We thus deduce that there is a local maximum in the refractive index just below the absorption line with (nmax − n) = 0.027. If the exciton absorption is saturated, this local maximum will disappear. Hence the maximum change in the refractive index is 0.027. (11.16) (a) The saturation density for the excitons is equal to the Mott density given in eqn 4.8. We consider here only the n = 1 exciton, since this is the ground state of the system and has the highest stability. 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 ). Equation 4.2 then gives r1 = 12.5 aH /0.06 = 11 nm, and we hence find NMott ≈ 1.8 × 1023 m−3 . (b) We first calculate the energy of the n = 1 exciton. We see from eqn 4.4 that the n = 1 exciton absorption line will occur at hν = Eg − RX , where Eg is given in Table D.2 as 1.42 eV. With µ = 0.06m0 and ²r = 12.5, we find from eqn 4.1 that RX = (0.06/12.52 )RH = 5.2 meV. The exciton energy is thus 1.41 eV at low temperatures. (We consider low temperatures here because the exciton would be ionized at room temperature.) The saturation intensity Is is the optical intensity required to produce the Mott density worked out in part (a). By using the result of Exercise 5.6(b), namely: Iατ N= , hν we then find Is =

hν 1.41 eV NMott = 6 × (1.8 × 1023 ) ≈ 4 × 107 W m−2 . ατ 10 · 10−9

117


Optical Properties of Solids 2nd Ed by Mark Fox

Recommend Documents