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of the em wave is: < p >= -s < E2 > + - / / < H2 > 2 2
(5)
For a plane em wave Eq.(5) becomes:
=e<E2> The substitution of (6) into (1), gives:
(6)
ANSWERS
24
=-nl
(7)
For 7=100 W/m2 from (7) we obtain: >=3.3xKT 7 J / m 3
(8)
Note: Consider a surface S perpendicular to the direction of propagation of an em plane wave (see Fig. 2.1). The energy, E, flowing through S in the time interval At is equal to the energy contained in a cylinder with base S and height l=cAt/n. Since E = / S At, the energy density is given by: p = E/(Sl) = nIIc, as obtained above.
t= Fig. 2.1 Energy density of an em wave
2.2A Photon flux of a plane monochromatic wave. The photon flux of a plane monochromatic wave of intensity / is given by:
hv he If A=500 nm, we get: F=5x 1020 ph m"2 s 1 If A=100 //m, then: F=lxlO 23 ph m"2 s"1.
2.3A Number of modes of a blackbody cavity. The number of modes per unit volume and per unit frequency range is given by (2.2.16) of PL: 1 dN Pv
~Vdv~
STTV2
(1)
25
2. INTERACTION OF RADIATION...
where we have considered an empty cavity (w=l). The number of modes for a cavity volume Fthat fall within a frequency bandwidth A vis given by: N = pvV\Av\
(2)
where we assumepv constant over Av. To solve the problem we have tofindthe relationship between Av and AX. Since Xv=c, we have: Av = ~AX
(3)
Using (1) and (3) in Eq.(2) wefinallyobtain: N = ^-VAX = 1.9xlOn
(4)
Notes: (i) Another way to solve the problem is to calculate the number of modes per unit volume and per unit wavelength range, /u, which is related to pv by the following relation: pxdX = -pvdv
(5)
where we use the minus sign because dX and dv have opposite signs, while px andp^are both positive. From (5) we get:
Px
Pv
dv zz Pv
^~ Hx
c
H*
Replacing p v in (6) by the expression given in (1), we obtain:
The number of modes falling within the bandwidth AX for a cavity volume V is therefore given by: ^
(8)
At
(ii) Expression (3) can be rewritten in the following compact and useful way: ^ — ^ v X
m
26
ANSWERS
2.4A Wien's law. We define px according to the relationship : = -pvdv
(1)
where we use the minus sign because dA and dv have opposite signs, while px and /cvare both positive. Since v=c/A, we have: — =~
(2)
from (1) and (2) we get: _
dv _ c
Px= Pv
=
~ 'd~X
~x}Pv
Using the expression (2.2.22) of PL for pn we obtain: Jnhc A
1
^
5
To find tlie maximum of px , we first calculate dpA/dA and equate it to zero. We get:
Taking y-(hcl
( he ) 1 /7C f he ) exp -1 = exp \AkT) J JLkT \AkT) AkT), Eq.(5) can be rewritten as: 5[l-exp(-v)] = y
(5)
(6)
This is a trascendental equation, which can be solved by successive approximations. Since we expect that (heIAkT)-hvIkT»\, we also expect that exp(-y)«l. As a first guess we can then assume y=5. Substituting this value in the left hand side of Eq.(6) we obtainy=4.966. Proceeding in the same way we rapidly obtain the solution of Eq.(6):y=4.9651. We thus obtain: hc
kTAm
=4.9651
(7)
which can be written in the following form: (8)
27
2. INTERACTION OF RADIATION...
which is the Wien's law. Note: The iterative procedure to find the point of intersection between two functions, converges if we start with the function with the smaller slope near the point of intersection. Consider, for example, two functions fix) and g(x) (see Fig. (2.2)). g(x) has the smaller slope near the intersection point. First we calculate g(x0), where JC0 is the guessed solution. Then we solve the equation./(jr) = g(x0) and we find the next order solution x=x}.Then we calculate g(x\) and we proceed as before. In this way we generally have a fast convergence of the solution.
Fig. 2.2 Iterative procedure to find the intersection between two functions.
2.5A Blackbody cavity filled with a dispersive medium. The number of modes per unit volume and per unit frequency range (radiation mode density) is given by (2.2.16) of PL:
1 dN{y) V dv
0)
where N(v) is the number of resonant modes with frequency between 0 and v. According to (2.2.15) of PL it is given by:
28
ANSWERS
()
\ 3c3n
()
3c3
where c is the velocity of light in vacuum and n = n(v) is the refractive index of the dispersive medium filling the blackbody cavity. Using Eq.(2) in (1), we obtain:
dv1
—
C3
„ „ „ + „ _
^
dv)
c
(3)
Since
dn dX dn _ c dn _ X dn 7^~~dv7x~~'^'dX~'~~c~dX tlie group index ng can also be expressed as follows: dn cf n =n + v— = /? + — g dv X[
X2 dn)
. dn \ = n-X— c dX) dX
... (5)
Notes: i) If the refractive index does not significantly vary with wavelength in a given spectral region, we can assume ng = n. ii) Material dispersion is due to the wavelength dependence of the refractive index of the material. Given a homogeneous medium, such as a piece of glass or an optical fiber, characterized by a dispersion relation /?= p(co), the group velocity is given by ug=do)/dp
= (dfi/d(Q)~\
where is the wavelength (in vacuum) of an e.m. wave whose frequency corresponds to the center of the line.
2. INTERACTION OF RADIATION...
35
Equation (3) can be used either to obtain the value of cr, when rsp is known, or the value of rsp when <jis known. If o(v) is known, to calculate rsp from (3) we multiply both sides by dv and integrate. Since \gtiy ~vo)dv = 1, we get:
2.13A Radiative lifetime and quantum yield of the ruby laser transition. From Eq.(3) in Problem 2.12 we have:
For a Lorentzian lineshape at v = v0 we have g(0)=(2/7rAv0). For ^o = 694 nm, we obtain from Eq.(l) the spontaneous emission lifetime (i.e., the radiative lifetime)
rsp = -^- — = 4.78 ms KnrrnAvQG
(2)
According to (2.6.22) of PL, tlie fluorescence quantum yield is given by: (j> = — = 0.625
(3)
2.14A Radiative lifetime of the NdrYAG laser transition. The number of atoms decaying radiatively per unit volume and unit time on the actual laser transition is given by:
Where N22 is the population of the sublevel m=2 of the upper laser state, and rr>22 is the radiative lifetime of the R2 -»r 3 transition. The total number of atoms decaying radiatively from the two sublevels of the upper state per unit volume and unit time is:
36
ANSWERS
N2
r2=~T
(2)
r,2
where: N2 and rr,2 are the total population and the effective radiative lifetime of the upper level, respectively. Since the ratio between (1) and (2) is equal to 0.135, rr,22 is given by: r
_N22
r,2
f
where f22 represents the fraction of the total population found in the sublevel m=2. Since die two sublevels are each doubly degenerate, then, according to Eq. (2.7.3) of PL, one has:
—I
(4)
kT) where AE is the energy separation between the two sublevels. We then obtain:
y%
'^ x
22 =
22 T T
rr~
:
^21+^22
For AE = 84 cm"1 and kT = 208 cm"1 (T = 300 K), we obtain f22 = 0.4. The radiative lifetime rry2 is given by:
where is the quantum yield. Substituting the calculated values off22 and rr>2 in Eq.(3) we finally obtain: fr,22= 1.2 ms
2.15A Transient response of a two-level system to an applied signal. We consider a two level system, and we assume that the two levels have the same degeneracy (g\-gi)> We suppose that at time /=0 the population difference AN(0)=N\(0)-N2(0) is different from the thermal equilibrium value, zLVe=Arie-Ar2e. A monochromatic em wave with constant intensity / is then
37
2. INTERACTION OF RADIATION...
turned on at t=O. The rate of change of the upper state population N2 due to the combined effects of absorption, stimulated emission and spontaneous decay (radiative and nonradiative) can be written as: dNj ,.,/„ __ \ No—N% v l x/ dt x where W=ollhv. The term describing the spontaneous decay takes explicitly into account that, without any applied external signal (i.e., for 7=0), the population density N2 relaxes toward the thermal equilibrium value N2. We then indicate by Nt the total population density and by AN the population difference:
Nt=Nx+N2
(2)
AN = NX-N2
(3)
Using (2) and (3) in (1) we get:
AN-\Nt-2N2) r
dt
(4)
Since Eqs.(2) and (3) are valid also at thermal equilibrium, we obtain: ANe = Nf -N\ = Nt-2N\
(5)
From (4) and (5) we obtain the final form of the rate equation for the population difference AN:
(6)
dt
which can be easily solved, by variable separation, with the initial condition: AN= AN(0) at t=0. We obtain: iMV
exp( I
1 + 21
M. I +.WW t,
l"~
I
/*7\
From (7) we see that, with no applied signal (i.e., when 7=0, and hence W=0), the population difference AN(f) relaxes from the initial value AN(0) toward the thermal equilibrium value AN6 with the exponential time constant r. In the presence of an em wave of constant intensity 7, the population difference AN(t) is driven toward a steady-state value AN^ given by:
ANSWERS
38
^00 =
ANe
ANe hv
ANe -I
(8)
1+— Is
where Is = (hvllar) is the saturation intensity. The time constant, z1, by which this equilibrium is reached is seen from Eq.(7) to be given by: r 1+ //A
(9)
and it decreases upon increasing /. Note that, from Eq.(8), it is apparent that AN^Abf. Using (8) and (9), Eq.(7) can be rewritten in the following more compact form: ( t\ AN(t)=ANo0-h[AN(0)-AN(X))cxp\ (10) Figure 2.3 shows the temporal behavior of AN(t) for different values of the normalized intensity, I/Is.
AN(t)
Fig. 2.3 Temporal evolution of the population difference AN(t) for different values of the normalized intensity 1/IS.
39
2. INTERACTION OF RADIATION..
2.16A Gain saturation intensity. We consider the case where the transition 2->l exhibits net gain. We assume that the medium behaves as a four-level system (see Fig.2.4), and the inversion between levels 2 and 1 is produced by some suitable pumping process.
N2/T2g
WnN\
Fig. 2.4 Energy levels and transitions involved in gain saturation of a four-level laser. We further assume that transition 3-*2 is so rapid that we can set A^=0. We denote by \IT2\ and l/z^ the decay rates (radiative and nonradiative) of the transitions 2->l and 2-»g, respectively, where g denotes the ground state. The total decay rate of level 2 is thus given by: 1 T
2
1 T
2\
1 T
(1)
2g
With these assumptions, we can write the following rate equations for the populations of levels I and 2:
(2)
where Rp is the pumping rate. In Eq.(2) we have [Eq.(l. 1.8) of PL]:
(3) Si
40
ANSWERS
where: g\ andg2 are the degeneracies of the levels 1 and 2, respectively. Eqs.(2) can thus be written as follows:
(4)
In the steady state (i.e., for dN]/dt=dN2/dP=O) Eqs.(4) reduce to a simple system:
Using Cramer's rule to solve this system we get:
R where A is the determinant of the matrix of coefficients of the system (5): 1
From (6) the population inversion can be written as follows:
(8)
Since ^21=021 ^"= 021 //Av, Eq.(8) can be rewritten as follows:
Is
41
2. INTERACTION OF RADIATION...
where AN0=\N2-—Nl
= RpT2\ 1-&--H-
is the population inversion
in the absence of the saturating beam (i.e., for 7=0) and: in/
i (10)
is the gain saturation intensity [in (10) 0=021]. From (10) it is apparent that if T\«r2, or T2K r2i, org2«g\ or any combination of the previous conditions, then the saturation intensity can be approximated as follows: l^
(ID
ar2
2.11 A Population inversion of a homogeneously broadened laser transition As shown in Problem 2.12, the stimulated emission cross section o(v) is given by: ji2l
8;r/r rsp %nn The cross section at line center is thus given by: )
(2)
For a Lorentzian lineshape (homogeneous broadening) at zJv^O we have [see (2.4.9b) of PL]: g(0) = - ^ -
(3)
Assuming n=l we get: a(0) = A2i
£
2
=9.68xl0~ 22 m2
(4)
42
ANSWERS
The gain coefficient at line center is given by (2.7.14) of PL:
1 = 5 nT1
(5)
The population inversion is thus given by: ^ 2 - ^ 1 — = —= 5.16xl0 2 1 nT 3
(6)
The gain saturation intensity is given by (11) in Problem 2.16. Since T2»T\ and g\»g2 we can use Eq.(12) of Problem 2.16: (7)
ar2
2.18A Strongly coupled levels. To prove Eq.(2.7.16a) of PL we assume that the upper level, 2, consists of g2 sublevels with different energies but with very rapid relaxation among them. Each sublevel, 2/ (/=1,2,..., g2), may also consist of many degenerate levels, 2jk (£=l,2,...,g2/, where g2j is the degeneracy of sublevel 2/). Due to the rapid relaxation, Boltzmann statistics can be taken to hold for the population of each individual level. We can then write: CD where Nyk is the population of the degenerate level k of the sublevel Ij. Since the degenerate levels of a single sublevel are also in thermal equilibrium, their population must be all equal, thus: N2Jk = - ^
(2a)
where N2J is the population of the sublevel 2/. Similarly, for the first sublevel, 21, we have: ^1,
^ £21
From (2) and (1) we then obtain:
(2b)
2. INTERACTION OF RADIATION...
""^•irT^^1
43
Eg. Use the results to calculate the upper and lower levels in GaAs for a transition at 1.45 eV [effective masses in GaAs are mc = 0.067 AW0, mu = 0.46 m0, while bandgap energy is Eg = 1.424 eVJ.
3.15P Frequency dependence of the gain of an inverted semiconductor. Consider an inverted bulk semiconductor. a) Give the analytical expression of the gain as a function of photon energy at T = 0 K and find the energy for which the gain is the highest; b) explain qualitatively how these results are expected to be modified at room temperature.
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
51
3.16P Gain and gain bandwidth calculation in a GaAs amplifier. Consider a bulk GaAs semiconductor at room temperature and assume the following expression for the absorption coefficient: a
o ( v ) = 19760 \hv -Eg f
2
, where ao is expressed in cm"1 and the photon
energy hvin eV. Calculate the gain coefficient at a photon energy exceeding the bandgap by 10 meV and for a carrier injection N = 2xlO18 cm'3 (hint: use Fig. 3.15(b) in PL to determine the energies of the quasi-Fermi levels). Calculate also the gain bandwidth. Compare the results with those that would have been obtained had the semiconductor been cooled to a temperature approaching 0 K.
3.17P Differential gain of a GaAs amplifier. For a carrier injection of N = 2xlO18 cm'3 and a photon energy exceeding the bandgap by 10 meV, the gain coefficient of GaAs can be calculated to be g = 217 cm"1. Assuming a transparency density of Ntr = 1.2xlO18 cm'3, calculate the differential gain.
3.18P Thickness of a quantum well: an order of magnitude estimate. Consider a layer of GaAs of thickness L sandwiched between two AlGaAs barriers at room temperature (3T= 300 K). Estimate the layer thickness for which quantum confinement effects start to play a role for electrons in the conduction band (effective mass for electrons in the conduction band in GaAs is mc = 0.067 /wo). [Hint: calculate the De Broglie wavelength for thermalized electrons]
3.19P An ideal quantum well. Consider a particle of mass m in a one-dimensional potential well of thickness L, with infinite potential barriers at the boundaries. Using basic quantum mechanics calculate the discrete energy levels inside the well.
52
PROBLEMS
3.20P Energies of the quasi-Fermi levels in a semiconductor quantum well. Consider a semiconductor quantum well under non-equilibrium conditions with an injected carrier density Ne = Nh = N. Show in detail how to calculate the energies of the quasi-Fermi levels in the conduction and valence bands, respectively. (Level of difficulty higher than average)
3.2IP Calculation of the gain bandwidth in a GaAs quantum well. For a 10-nm GaAs quantum well at room temperature (T = 300 K) calculate (using Fig. 3.26 in PL) the overall bandwidth of the gain curve for an injected carrier density of N = 2x 1018 cm"3.
3. E N E R G Y L E V E L S , R A D I A T I V E A N D N O N R A D I A T I V E . . .
53
ANSWERS 3.1 A Vibrational frequency of a diatomic molecule. Let us consider the two atoms as two masses bound by a spring of elastic constant h- Projecting the equations of motion on an x axis, we get (la)
dr M2 —-^- = -k0(x2 -Xi) dt2
(lb)
Upon dividing Eqs. (la) and (lb) by Mx and M2, respectively, and then subtracting the resulting two equations, we get 9~V*2 "~ x \ ) ~ " " ^ 0 » y
T7
V2 ~
x
l/
v^/
Making the substitution^ = x2 - x\ and defining the reduced massMr as Mr
Mi
M2
we get the equation
42 +^ dt
=0
(4)
Mr
This is the equation of a harmonic oscillator of mass Mr and elastic constant k0. Its vibrational frequency is then given by (5) For the particular case of a homonuclear diatomic molecule (M\ vibrational frequency becomes
\l l7t\M
=M2=M)the
(6)
54
ANSWERS
3.2A Calculation of the elastic constant of a molecule. Let us first express the vibrational frequency in Hz. Recalling that v=v/c
= l/A
(1)
we obtain v = cv = 213cm'1 3 x 1010 cm/s = 6.4x 1012 Hz
(2)
Using Eq. (6) of the previous problem, which applies for a homonuclear diatomic molecule, we obtain the elastic constant of the molecule k0 = 4TT2V2 — = 170 —
2
(3)
m
3.3A From the potential energy to the vibrational frequency. Let Ro be the equilibrium internuclear distance of the molecule, corresponding to a minimum of the electronic energy. From a second order Taylor expansion of the energy around Ro, we get r
{\ (/?-*0)+Ip-£| (R-Roy+...
If Ro is a minimum position for the energy, we have (dU/dR)R
(i)
= 0, so from
Eq. (1) we get
+ ±(^f\
(R-R0f
(2)
Since this expression also gives the potential energy of the oscillator, the restoring force can be calculated as
and
is 2
elastic, 2
k0 = yi u/dR )p
i.e.
proportional
to
displacement,
with
a
constant
. Therefore the vibrational frequency of the molecule is:
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
55
(4)
M
and can thus be directly calculated from the electronic energy function, known from experiments orfromab initio calculations. Note: Any potential energy function, in the neighbourhood of the stable equilibrium position, can be approximated by a parabola: this explains the importance of the harmonic oscillator in physics.
3.4A The Morse potential energy. It is a simple matter to show that Ro is a minimum position for the Morse potential energy and that the corresponding energy value is U(R0) = 0. For large values of the internuclear distance, the Morse potential tends to its asymptotic value U(<x>) = De ; therefore the dissociation energy, i.e. the energy that must be delivered to the molecule to bring its nuclei far apart, is simply given by U(n)-U(R0)
= De
(1)
To calculate the vibrational frequency, we can perform a Taylor expansion of the Morse function around 7?0, as in the previous problem; more simply, we can expand the exponential function to the first order around Ro, getting (2) We can thus express the Morse potential around the equilibrium position as U(R) = Dep2{R-R0)2
=-ko(R-Ro)2
(3)
with k0 = 2De/3 . The expression of the vibrational frequency for the molecule then follows from Eq. (6) of problem 3.1:
i
Un./i2 (4)
ANSWERS
56 3.5A Calculation of the Franck-Condon factor.
Assuming potential energies for ground and excited state with the same curvature, we can write the wavefunctions of the lowest vibrational levels in ground and excited states as
Vog=\
1
.1/2
an
y
(la)
exp - -
(lb)
exp -
1/2 All
an
1/2
with a2 = h/(mk)l/2 , y = ( f t - ^ J / a , / = (R-ROe)/a , #og the equilibrium internuclear separations in the ground and respectively. It is for us more convenient to write y' = A = \ROe - ROg )ja . To calculate the Franck-Condon factor,
and /?Oe being excited state, y - A, with we need to
evaluate the overlap integral (2) an By some simple mathematical manipulations, this expression can be put in the form
^oo=-~-exp[-A2/4Jjexp - I J ' - y l
/2
dy
(3)
By making the additional substitution g = y -A/2 and recalling the mathematical result given in the text of the problem, we obtain 2
= exp [- A
/4j= exp
~ R0g f 4a2
(4)
The Franck-Condon factor for the given vibronic transition is then given by 2 = exp
~ Rog f 2a'
(5)
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
57
Note that the Franck-Condon factor rapidly decreases for increasing difference in equilibrium internuclear distances between ground and excited state.
3.6A Rotational constant of a diatomic molecule. Let us first calculate the position of the center of mass of the molecule. Assuming an x axis oriented along the direction joining the two nuclei and with the origin on the first atom, we get M2R0
0 = x cm
A/j +A/ 2 Mx +M2 The moment of inertia about an axis passing through the center of mass and perpendicular to the internuclear axis is then given by l=M{x2cm+M2(R0-xcm)2
(2)
By inserting Eq. (1) into Eq. (2) and by some simple algebraic manipulations, we obtain the expression
Rl=MrRl
(3)
where Mr = MxM2/(Mi +A/ 2 )is the reduced mass of the molecule as previously defined in problem 3.1. According to classical mechanics, the kinetic energy of a rigid body rotating around a given axis can be written as Ek - L2/2I, where L is the angular momentum and / is the moment of inertia about that axis. According to quantum mechanics, the angular momentum is quantized, i.e. it can have only discrete values, given by the quantization rule I? = h2j(j + l), where J is a positive integer. By substituting into the preceding expression for kinetic energy, we obtain
(4, with the rotational constant B given by
2/
%
58
ANSWERS
Note: For spectroscopic purposes, it is often convenient to express the rotational constant in inverse wavenumbers *
(6)
^
3.7A Far-infrared absorption spectrum of an HC1 molecule. Let us first calculate the energy spacing between two consecutive rotational levels of rotational quantum numbers J-l and./: AE = E(j)~E(J
- l ) = B J(J + l ) - B (j -\)j
= 2BJ
(1)
We thus see that the energy difference between two consecutive rotational levels is not constant, but increases linearly with increasing quantum number J. This property proves to be consistent with the measured far infrared absorption spectra of the HC1 molecule. We can in fact extract nearly the same rotational constant from the different transitions, namely J = 3 - > J = 4 5 = AE/8= 10.41 cm1 J= 4_
>
J=5
B = AE/10 = 10.4
cm1
J = 5 - > J = 6 £ = AE/12= 10.39 a n 1 We can thus assume for the rotational constant of the molecule the value B = 10.4 cm"1. To determine the interatomic equilibrium distance starting from the rotational constant, let us first calculate the reduced mass of the molecule: = 1.624 xKT 2 7 r
m
kg
(2)
m
H + Cl The equilibrium interatomic distance can then be obtained from Eq. (6) of the previous problem as Rn = 1 0
^—^7=0.128
nm
(3)
\AMB
Note: The experimental data show a slight decrease of B for higher energy levels; this can be understood in terms of centrifugal effects. For the higher energy levels,
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
59
as the molecule is spinning more rapidly, the interatomic bond is elongated slightly, causing an increase of the moment of inertia and a corresponding decrease of B .
3.8A The most heavily populated rotational level. It was shown in problem 3.6 that, in quantum mechanics, the angular momentum of a rotator is quantized according to the relationship: L2=h2j(j
+ \)
(1)
where J is an integer. Consequently, also its energy is quantized: 1)
(2)
However, also the direction of angular momentum is quantized; its projection on an arbitrary axis z can in fact assume the 27+1 different values Lz=mh
m = 0,±l,...,±J
(3)
Therefore a rotational level of quantum number J is (2y+l)-fold degenerate. The probability of occupation of a given rotational level can then be written as p(j) = C(2J + l)exp[- Rl(j + \)lkT)
(4)
where C is a suitable normalization constant. To calculate the most heavily populated level, we equal to zero the derivative of/? with respect to J:
^- = Cexp[- BJ(J + l)/£r]{2 - (2J + \)2BlkT]= 0
(5)
cLJ This equation holds for a rotational quantum number
It is easy to verify that this corresponds to a maximum for the level occupation probability. Taking the IC1 molecule (B = 0.114 cm 1 ) and recalling that at room temperature (T = 300 K) kT = 209 cm"1, we obtain J^ = 29.8. This means that the level with rotational quantum number J = 30 is the most heavily populated for a IC1 molecule at room temperature.
60
ANSWERS
3.9A The emission lines of a CO2 molecule. In rotational-vibrational transitions for most diatomic and triatomic molecules (and in particular for CO2), selection rules require that the rotational quantum number is changed by one unit: A/=JM-J'=±1
(1)
f
where J" and J arc the rotational numbers for the lower and upper vibrational states respectively. For the so called "P -branch" transitions we have AJ = +1, i.e. the rotational number of the lower vibrational state is higher, J" = J' +1; this is referred to as the P(J") transition. It we let hv0 be the energy difference between the two vibrational levels, the energy of the P(J") transition can be easily calculated as = hv0 + BJ'iJ'+i)- BJ"(j"+l)=
hv0 -2BJ"
(2)
In particular, taking the energy difference between two P-branch transitions with rotational numbers J'\ and J\ we get E(J%\)-E(J"2)=2B{J"2-J%\)
(3)
Eq. (3) can be used to calculate the rotational constant of the molecule. For our problem, let us first express the energies of the two transitions in cm'1 [we recall that E (cm1) = MX (cm) ]. We obtain £(12) = 951.16 cm'1, £(38) = 930.92 cm 1 . With the help of Eq. (3), we obtain B = 0.389 cm"'. The energy difference hv0 between the two vibrational levels can then be obtained from Eq. (2): hv0 =E(J") + 2BJ"
(4)
Using tlie previously obtained value of B, one easily calculates hvo= 960.5 cm'1.
3.I0A The law of mass action. Let us write the expression of the electron density in the conduction band at thermal equilibrium, using the system of coordinates of Fig. 3.9a in PL (i.e. measuring energy in the conduction band from the bottom of the band upwards): 00 N€=\pc(Ec)fc{Ec)dEc o with the density of states given by
(1)
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
61
and the level occupation probability given by the Fermi-Dirac distribution:
In our frame of reference one has EF < 0, since the Fermi level is lying in the bandgap; if we additionally assume EF » kT, we see that for any value of Ec tlie exponential function in tlie denominator will be much larger than 1. Therefore we can write: fc{Ec)*exp[-(Ec-EF)lkT]
(4)
which corresponds to using a Boltzmann approximation for the tail of the Fermi-Dirac distribution. By inserting (2) and (4) into (1), we get: (l ) 3 / 2 °° Ne = V ^ 3 exp(g F /KT) J ^ / 2 exp(- EjkT)dEc 2/r h 0
(5)
By making the change of variables x = Ec/kT and recalling tlie mathematical result given in the text of tlie problem, we can easily obtain: e
2~
v h Upon interclianging tlie index c with u, tlie same calculations can be repeated to obtain the hole density in tlie valence band. Tlie only difference is that, in tlie coordinate system used in Fig. 3.9a for tlie valence band, the energy of the Fermi level is given by - Eg - EF. Therefore one obtains tlie hole density as (7) The product of electron and hole concentrations is then given by:
NeNh = 4^f\
{mcmv ) 3/2 exp(- Eg /ICT)
(8)
This product is independent of the position of the Fermi level within the bandgap, i.e. of the doping level of the semiconductor, and it applies as long as
62
ANSWERS
the system is at thermal equilibrium. This important result of semiconductor physics is referred to as the "law of mass action".
3.11 A Energies of the quasi-Fermi levels. Let us calculate the electron density in the conduction band:
oo Ne=jpc(Ec)fc{Ec)dEc o
(1)
with the density of states given by Pc\Ec)-— 2
In I h2
For the ideal case of a semiconductor at T = 0 K, the Fermi-Dirac distribution function simplifies and becomes
Ec<EFc Ec>EFc Therefore expression (1) simplifies to
The energy of the quasi-Fermi level, with respect to the conduction band edge, is then readily obtained from (4) as:
EFc=ty*[
^-^e
(*)
Upon interchanging the indexes c with u and e with h, an analogous expression can be obtained for the energy of the quasi-Fermi level in the valence band, calculated with respect to the valence band edge:
Eqs. (5) and (6) show that, by increasing the injection density, the quasi-Fermi level move deeper in to the conduction and valence bands. They are rigorously
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
63
valid only at T = 0 K, however they hold approximate validity at low temperatures or high injection levels, when EFCiU » kT and the Fermi-Dirac distribution can be approximated reasonably well by a step function.
3.12A The quasi-Fermi levels in GaAs. The energy of the quasi-Fermi level in the conduction band can be calculated from the expression derived in the previous problem upon inserting the values of the constants relevant to our case: EFc = G;T j
-^—Ne 2m c
= 86.4
meV
(1)
The quasi-Fermi energy in the valence band can then be calculated as: EFv =
Y ^-EFc = 12.58 mv
meV
(2)
To evaluate whether the low-temperature approximation still holds at room temperature, we need to compare the above energies with kT (« 26 meV at T = 300 K). Since EFc » kT , i.e. the quasi-Fermi level lies well within the conduction band, we expect the low-temperature approximation to hold reasonably well at room temperature; on the other hand, since EFu < kT, the valence band approximation is expected to be inaccurate in this case. In fact, from an inspection of Fig. 3.15(b), which plots the results of exact calculations at room temperature, we see that, at an injected carrier density of 2xlO 18 cm"3, one has EFc = 3.2 kT» 83.2 meV, while EFv = - kT*-26 meV. This confirms the validity of the low-temperature approximation for the conduction band but not for the valence band.
3.13A Derivation of the Bernard-Duraffourg condition. To have net gain in a semiconductor at a given frequency v 0 = (E2 *-EX ')//?, the number of transitions available for stimulated emission must exceed the number of those available for absorption. According to Eq. (3.2.31) in PL, the number of transitions available for stimulated emission is given by,
64
ANSWERS
where dN is the overall number of transitions between frequencies v0 and v0 + dvo, fc(E2%) is the probability that the upper level is full and \-fJiE\) is the probability that the lower level is empty. The number of transitions available for absorption is, according to Eq. 3.2.30 in PL dNa=dNfv{Ex>\\-fc{E2>)}
(2)
where, in this case,/u(£i') is the probability that the lower level is full and 1f)-fo{Ex>)}
(3)
Having net gain therefore requires that: fc(E2')> fv(El')
(4)
or, using Eqs. (3.2.11) in PL:
>
I
1
(5)
1 + exp[(£ 2 '-£ Fc >)lkT\ 1 + Qxp[(El'-EFu')/kT] which becomes Ei'-EFo'>
E2'-EFc'
(6)
or, equivalently E^-E^E^-E^
(7)
which is the sought condition.
3.14A Laser levels in a semiconductor. Within the parabolic band approximation, the energies of electrons and holes in the conduction and valence bands are given by h2k2 E2=^
(la) 2
3. E N E R G Y L E V E L S , R A D I A T I V E A N D N O N R A D I A T I V E . . .
65
We recall once again that the energy of the conduction band is measured from the bottom of the band upwards, while the energy of the valence band is measured from the top of the band downward. Since the wave vector of a photon is negligible with respect to that of electron and holes, selection rules require that optical transition occur vertically in the E vs. k diagram, i.e. that the wave vector is conserved: kc=kv = k (2) The energy of the optical transition can thus be written as 2 2
k ( 1
^ \
h2k2
\ \
\
g
?^ 2mr
(3)
where mr is the so-called reduced mass of the semiconductor, given by (4) From Eq. (3) the k value corresponding to a given transition energy hv0 is readily calculated. Upon substituting the resulting expression into Eqs. (1), we obtain, in the usual frames of reference, the energies of the upper and lower laser levels (5a) (5b) To apply these results to the case of GaAs, we first note that the reduced mass is given by: Ytly. — mv +mc
— U.lOoD /WQ
For Eg = 1.424 eV and hv0 = 1.45 eV, we then obtain: E2 = 22.3 meV, Ex = 3.3 meV.
3.15A Frequency dependence of the gain in an inverted semiconductor. According to Eq. (3.2.37) in PL, the gain coefficient of an inverted semiconductor is given by
(p)
66
ANSWERS
(1) with (2) 1
l + exp[(E2-EFc)/kT] 1 ^expl(E[-EFo)/kT)
(3) (4)
At T= 0 K the Fermi distribution functions simplify considerably to become (5)
(6) We then get (7) The gain coefficient can then be written as
aQ(y)
~Y vmax the gain changes abruptly sign and becomes absorption. In the case of a room temperature semiconductor, the Fermi distribution function shows a smoother transition from 1 to 0 and also the abrupt frequency jump in the gain function disappears.
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
67
3.16A Gain calculation in a GaAs amplifier. Let us first calculate the values of the energy for the two levels involved in laser action for a transition energy exceeding the bandgap by 10 meV. According to the results of problem 3.14, the energy of the upper level in the conduction band is
J^
(1)
^ ^ 0.067 m0
while that of the lower level in the valence band is E l=(f,y-EJ^ 1
Y
iomeV^^Ul.2meV
=
*'m
u
(2)
0.46m0
The optical gain of the semiconductor is then given by g(y) = cc0(y)[fc(E2)-fo(El)] where the absorption coefficient is ao(v)= 19760 (hv-Eg}/2 =1976 cm"1
(3) (4)
To compute the Fermi occupation factors, the energies of the quasi-Fermi levels in the valence and conduction bands need to be known. From Fig. 3.15(b) in PL we obtain, for a carrier injection JV = 2x 1018 cm'3: EFc = 3kT=lS
meV
EFu = -1.5 JeT= -39 meV
(5)
The Fermi occupation factors can then be obtained as
fc{Ei) =
= 0.934
(6)
fv(Ex) =
= 0.824
(7)
Inserting the above values in Eq. (3), we obtain the gain coefficient as g = 217 cm"1. If we consider the semiconductor at 0 K, the energies of the quasi-Fermi levels at a carrier injection density N = 2x 1018 cm"3 can be calculated using the results of problem 3.11:
EFc = — [to2tf p = 86.4 meV 2.TYI ~
(8)
68
ANSWERS
EFv=^EFc mu
= 12.58 meV
(9)
Note that the energy of the quasi-Fermi level of the conduction band, which was lying well within the band already at room temperature, did not change much going to 0 K, while the quasi-Fermi energy of the conduction band experienced a large change. Recalling the Fermi occupation factors at T = 0 K, which were given in the previous problem, we obtain:
/cfoH
/w(Ei) = 0
(10)
and therefore the gain coefficient becomes g=a0=l916
cm"1
(11)
Therefore, going to low temperatures, the gain of the semiconductor increases significantly. The gain bandwidth can be obtained from (3.2.39) of PL. Upon switching from the primed to the unprimed coordinate system, we write according to (3.2.3)
so that, from (3.2.39) of PL, we get Eg c -Eic)lkT)
(9)
In Eq. (9) the sum is extended over all subbands and we assume that the effective mass of the electrons does not vary in the different subbands. By interchanging the subscript c with u we get, analogously, the hole density in the valence band mvkT
*v-Eiv)l1
Equations (9) and (10) can be used to obtain two plots of Ne vs. EFc and Nh vs. EFv, respectively. From these two plots, the values of EFc and EFu, for a given value of the carrier injection N = Ne = Nh, can then be obtained.
3. ENERGY LEVELS, RADIATIVE AND NONRADIATIVE...
73
3.21 A Calculation of the gain bandwidth in a GaAs quantum well. The condition for net gain in a semiconductor quantum well can be written as [see (3.3.26) in PL] f
80
PROBLEMS
4.21P Focusing a gaussian beam inside a piece of glass. Consider a gaussian beam with spot size wox and plane wavefront entering a lens of focal length/(assume zm »J). A long block of glass with refractive index n is placed at a distance L 10 GHz. Leaving a safety margin, we can choose: AVFSR= 15 GHz. This corresponds to a mirror distance: 3xl0 11 mms" 1
c 1
=
,_ = 10 mm
2xl5xl0 9 s" 1 The resolving power of the interferometer is given by A vM = A VFSRIF, where F is the finesse of the instrument. To observe the single longitudinal modes, we need to have AVM < 100 MHz; again leaving some margin, we can choose AvM = 15 MHz, which corresponds to a finesse F = 200. Assuming to use two mirrors with the same reflectivity and using the result of the previous problem, we obtain the required mirror reflectivity: R = 1 - rdF - 0.984. Finally, to get the excursion of the piezotransducer we recall that, to cover a free spectral range of the interferometer, the mirror distance should be varied by half a wavelength, i.e. ALFSR =/l/2 = 0.532 nm. R
4.8A An imaging optical system. The matrix formulation of geometrical optics relates the position rx and the slope r\ of a ray of light at the input plane of an optical system to the corresponding position and slope at the output plane, according to:
r2=Arx+Br\
89
4. RAY AND WAVE PROPAGATION... If B = 0, we get:
r2=Ar\
(2)
This condition means that all the rays emerging from a point source in the input plane of the system located at a distance r\ from the axis, regardless of their slopes r\, will converge to a point in the output plane at a distance Ar\ from the axis: the system thus images the input plane onto the output plane with magnification^. We can verify this condition considering an optical system made of a propagation do, a thin lens of focal length / and a propagation
