(11.1) In the Bohr model for hydrogen, the radius of the electron in the nth quantum level is given by:
rn= 4π²0~2 me2
n2 Z = n2
ZaH.
The magnitude of the electric field is given by the standard Coulomb formula:
E = Ze 4π²0r2,
which, on inserting rn from the Bohr formula, gives:
E = Ze
4π²0r2n = e 4π²0a2H
Z3 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 × 1011V 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
A =Epulse/τpulse
πr2 = 1 ÷ 10−8
π(2.5 × 10−3)2 = 5.1 × 1012W/m2. Hence with n = 1 for air, we find from eqn A.44 that E = 6.2×107V m−1. (b) The optical intensity is found from:
I = P
A = 10−3
20 × 10−12 = 5 × 107W/m2.
Then with n = 1.45 as appropriate for the fibre, we find from eqn A.44 that E = 1.6 × 105V 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.
2Zeff 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.
(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 in-version 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 indi-cated in Fig. 11.2. (Spontaneous emission can be neglected if uν is suf-ficiently large.) The absorption and stimulated emission rates are equal to N1B12uνg(ν) and N2B21uνg(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 stimu-lated emission rate becomes increasingly significant. Eventually, we reach a stage where N1= N2= N0/2, and the stimulated emission and absorp-tion rates are identical. There is therefore no net absorpabsorp-tion 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 N2are:
dN1
dt = −B12N1uνg(ν) + B21N2uνg(ν) , dN2
dt = +B12N1uνg(ν) − B21N2uνg(ν) .
On setting B12= B21 as appropriate for non-degenerate levels, and sub-tracting, we find:
d
dt(N1− N2) =d∆N
dt = −2B12uνg(ν)∆N , where ∆N = N1− N2. Integration yields:
∆N (t) = ∆N (0) exp(−2B12uνg(ν)t) , which, with ∆N (0) = N0, gives the required result.
3This shows that it is not possible to achieve population inversion (i.e. N2 > N1) in a 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.
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.
direction of propagation
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:
Only Pz(2) 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 E0. It will then be the case that Ex = E0cos θ and Ey = E0sin θ, and hence that:
Pz(2)= 2d36E20cos θ sin θ = d36E20sin 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.
With the data given in the exercise, this gives:
(11.8) (a) In the absence of the field, the index ellipsoid given in eqn 11.54 is of the form:
x2+ y2 n2o +z2
n2e = 1 .
On substituting into eqn 11.60 with Ex = Ey = 0 and Ez = E, we find that the only non-zero changes to the index ellipsoid induced by the field are as follows:
From eqn 11.56 we then see that the modified index ellipsoid is:
µ 1 which can be written in the form :
x2+ y2
We assume that the field-induced changes are small and use the approxi-mation (1 + x)−1/2= (1 − x/2) for small x to obtain the final result:
no(E) = no−1
2n3or13E , ne(E) = ne−1
2n3er33E .
(b) The geometry of the crystal is shown in Fig. 44. If the light is prop-agating along the y axis, and is polarized along the z direction, then it
x y z L
V x
y z
x y z L
V
Figure 44: Experimental geometry of the electro–optic phase modulator consid-ered in Exercise 11.8.
will experience a refractive index of ne(E). From part (a) we know that ne(E) = ne− n3er33E/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
λ∆neL = −πn3er33EL/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
λ ∆noL = −πn3er13EL/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.
y z x x¢ y¢
L
V
y z x x¢ y¢
L
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 Ez 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-y0 plane. We resolve the
light polarization vector into its components along the x0 and y0 axes.
The phase shift induced by a refractive index change ∆n in a medium of length L is, in general, given by:
∆φ = 2π λ∆nL ,
where λ is the vacuum wavelength. On applying this to the two compo-nents along the x0 and y0 axes, we then have:
∆Φx0 = 2πL
λ ∆nx0 =πLn30r41 λ Ez
∆Φy0 = 2πL
λ ∆ny0 = −πLn30r41
λ Ez,
where ∆nx0 and ∆ny0 are the field-induced refractive index changes along the x0 and y0 axes as given in the exercise. The phase difference ∆Φ is thus given by (with Ez= V /L):
∆Φ = ∆Φx0− ∆Φy0 =2πLn30r41Ez
λ = 2πn30r41V
λ .
Note that the phase change is independent of the length in this longitudinal geometry.
(b) On setting ∆Φ = π, we find:
Vπ= λ 2n30r41.
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
λ n3or63(V /L)L = 2πn3or63V /λ .
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/2πn3or63= λ/4n3or63.
On inserting the values of no and r63for λ = 633 nm, we find V = 4.2 kV.
(11.11) In a third-order nonlinear medium, the change in the relative permit-tivity is given by (see eqn 11.69):
∆˜²r= ∆²1+ i∆²2= χ(3)E2.
We therefore have ∆²2 = Im(χ(3))E2 and hence that ∆²2 ∝ Im(χ(3))I because I ∝ E2. 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 .
A saturable absorber has an absorption coefficient that obeys eqn 11.37.
In the limit of small intensities, this is of the form:
α(I) = α0− α0I/Is= α0− ∆α ,
where ∆α = α2I and α2= α0/Is. We thus have an intensity 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 = (Ex, 0, 0) .
The third-order nonlinear polarization is given by eqn 11.11, and, with Ey= Ez= 0, the nonlinear polarization is of the form:
Px(3)
Py(3)
Pz(3)
= ²0E3x
χxxxx
χyxxx
χzxxx
.
However, we see from Table 11.6 that χyxxx= χzxxx= 0. Hence we find P = ²0χxxxxE3x(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:
∆Φnonlinear= 2π
λn2IL = π , which implies:
I = λ
2n2L = 1.55 × 10−6
2 · 2 × 10−20· 10 = 3.9 × 1012W m−2. The optical power to produce this intensity is given by:
P = IA = 3.9 × 1012× π(2.5 × 10−6)2= 76 W .
This is a large power level for a continuous-wave laser, but not for a 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:
Ppk= Epulse
tp
= Pav
f tp
.
On inserting typical values, namely Pav = 10 mW, f = 100 MHz, and tp= 1 ps, we find Ppk= 100 W.
(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 Egand (Eg+ EFc+ EvF) 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:
Nh= 1 3π2
µ2
~2
¶3/2
(m3/2hh + m3/2lh ) (EFv)3/2,
which gives EvF= 0.007 eV for m∗hh= 0.5m0and 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 2n²0m0γω0,
where κmaxis the extinction coefficient at the line centre, while in part (c) it is shown that:
nmax= µ
²∞+ N e2 2²0m0γω0
¶1/2 .
It then follows, with ²∞= n2, that:
nmax= n
³
1 +κmax
n
´1/2 ,
where nmax is the maximum value of the refractive index. If we assume, as is demonstrated below, that κmax¿ n, we then find:
nmax= n + κmax/2 .
Now we know from eqn 1.19 and the data given in the exercise that κmax= λαmax
4π =847 × 10−9· 8 × 105
4π = 0.054 .
4In a real experiment, the behaviour would be more complicated due to many-body effects such as band gap renormalization. This causes a shift of Eg∝ −N1/3, which reduces the blue shift of the absorption edge.
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 × 1023m−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 Egis given in Table D.2 as 1.42 eV. With µ = 0.06m0and 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 Isis the optical intensity required to produce the Mott density worked out in part (a). By using the result of Exercise 5.6(b), namely:
N = Iατ hν , we then find
Is= hν
ατ NMott= 1.41 eV
106· 10−9 × (1.8 × 1023) ≈ 4 × 107W m−2.