Solutions to the Exercies of C. J. Foot’s Atomic Physics Chenchao Zhao Department of Physics, Beijing Normal University, Beijing, China (Dated: June 28, 2011)
1
Early atomic physics Key formulas:
1 1 − n2 n′2 mH RH = R me + mH ~2 = 0.529 × 10−10 m a0 = (Ze2 /4πǫ0 )me Ze2 /4πǫ0 1 E=− ∝ Z2 2a0 n2 α2 ∆E = 2 E n e2 /4πǫ0 α= = 1/137 ~c f = R∞ ((Z − σK )2 − (Z − σL )2 ) c 1 ~ω 3 ρ(ω) = 2 3 π c exp ~ω/kT − 1 N1 ~ω N2 = exp − g2 g1 kT 4 2 2 e x0 ω P = 12πǫ0 c3 eB ΩL = 2me 6πǫ0 me c3 τ= e2 ω 2 1 =R λ
1
(1.1) (1.2) (1.3) (1.4) (1.5) (1.6) (1.7) (1.8) (1.9) (1.10) (1.11) (1.12)
1.1
Isotope shift
To find wavelengths of Balmer-α transitions, we set n = 2 and n′ = 3 as 36 in Eq (??) Then λ = 5R , hence 36 (1/RH − 1/RD ) 5 1 1 36me − = 5R∞ mH mD 18me = 0.18 nm ≈ 5mH R∞
λH − λD =
where mD ≈ 2mH .
1.2
The energy levels of one-electron atoms
Since (1.4), and let m, n be the quantum numbers of He+ and H, neglecting the isotope shifts, the energy levels in agreement are those with 1/n2 = 4/m2 , namely m = 2n. Those wavelengths should have the ratio RHe λH = λHe RH mHe (mH + me ) = mH (mHe + me ) 4mH (mH + me ) ≈ mH (4mH + me ) ≈ 1.00041 which is in accord with the data set, 1.00041.
1.3
Relativistic effects
With n = 4, Eq (1.5) gives λ E 4 = = 2 = 75076 ∆λ ∆E α for the fact that λ = ch/E and dλ = −ch dE/E 2
= −(ch/E) (dE/E)
= −λ (dE/E)
This corresponds to a grating of ∼ 105 grooves.
2
(1.13)
1.4
X-rays
Eq (1.7) reduces to f = 3cR∞ Z 2 /4 when Z ≫ σ. Therefore,
1.5
√
f ∝Z
X-rays
Since E = hf = 3/4(hcR∞ )Z 2 = 13.6 eV × 0.75Z 2 , then it predicts absorptions at around 6.4 keV and 6.9 keV.
1.6
X-ray experiments
See http://www.physics.ox.ac.uk/ history.asp?page=Exhibit10
1.7
Fine structure in X-ray transitions
Energy of the electron in the L-shell should be E = (82/2)2 × 13.6 = 22.9 keV and Eq (1.5) gives ∆E =
α2 E = E/75076 = 0.3 eV n2
But Kα transition means an energy of Eα = 10.2(Z − 1)2 eV = 66.9 keV then
1.8
∆E = 4.5 × 10−4 % Eα
Radiative life time
Eq (1.10) provides the power of dipole radiation which is rate of change of energy, for the circular motion, the power doubles since a circular motion can be decomposed into two linear oscillations, τ = E/P = ~ω
12πǫ0 c3 ∝ 1/(er)2 ω 3 2e2 r 2 ω 4
For photons of wavelength 650 nm, ω = 2.89 × 1015 rad/s; let r = a0 , the life time will be 2.7 × 10−7 s.
3
Table 1: Frequency shifts B [T] 3 × 10−5 1 ΩL [MHz] 2.6 8.8 × 104 ΩL /(1014 Hz) 2.6 × 10−8 8.8 × 10−4
1.9
Black-body radiation
Setting N2 = 0.1N, N1 = 0.9N, g1 = 1, g2 = 3, Eq (1.9) gives exp(~ω/kT ) = 27 provided the wavelength λ = 600 nm, ω = 2πc/λ = 3.13×1015 rad/s. Therefore, T = 7.23 × 103 K and Eq (1.8) gives the density ρ = 4.70 × 10−16 J s/m3
1.10
Zeeman effect
The Larmor frequency is given by Eq (1.11), and the Earth magnetic field is about 3 × 10−5 T, then the frequency shifts are listed in Table 1.
1.11
Relative intensities in the Zeeman effect
One circular motion can be decomposed into to two orthogonal sinusoidal motions. Let I denote the intensity of three eigen-oscillations of the electron. Then, we have • Along the magnetic field, only circularly polarized lights can be observed and the intensities are Iσ+ = Iσ− = I; • Perpendicular to the magnetic field, motion along z direction and projected horizontal motions can all be observed, and • 2Iσ+ = 2Iσ− = Iπ = I since the projected horizontal motions are only half of the circular motions. Therefore, (a) Total intensity perpendicular to the field is 2I; (b) Ratio of intensities received along to perpendicular to the field is 2I/2I = 1.
4
1.12
Bohr theory and the correspondence principle
Energy of hydrogen atoms takes the form 1 E = K + V = −K = − me v 2 2 But me Then E=−
v2 e2 /4πǫ0 = r r2
1 e2 /4πǫ0 , 2 r
dE =
e2 /4πǫ0 dr 2r 2
Hence
e2 /4πǫ0 ∆r 2r 2 ~ The angular frequency is also given by ω = ∆E/~ =
ω2 =
v2 e2 /4πǫ0 = r2 me r 3
Equating the two expressions of ω, we have s √ r~2 = 2 a0 r ∆r = 2 2 me e /4πǫ0 which is equivalent to
∆r = 2(a0 r)1/2 . ∆n Approximate the equation above by the corresponding differential equa√ tion, namely r ′ = 2 a0 r, and the solution turns out to be r = a0 n 2 .
1.13
(1.14)
Rydberg atoms
Eq (1.4), then dE e2 /4πǫ0 d (n−2 ) e2 /4πǫ0 =− = = Ry/n3 dn 2a0 dn a0 n 3 and for n = 50, ∆E = 1.1 × 10−4 eV. The radius of such atoms is around 2500a0 or 132 nm according to Eq (1.14).
5
2
The hydrogen atom
Key formulas Ylm (θ, φ) = (−1)m Z
s
(2l + 1) (l − m)! imφ m e Pl (cos θ) 4π(l + m)!
(2.1)
π
dθ eimθ e−inθ = 2πδmn cos mθ cos nθ dθ = πδmn sin mθ sin nθ −π Z π dθ sin mθ cos nθ = 0
(2.2)
−π Z π
(2.3) (2.4)
−π
Z
dΩ =
Z
2π
dφ
0
Z
π
sin θdθ =
Z
2π
dφ
0
0
Z
1
d(cos θ)
β (j(j + 1) − l(l + 1) − s(s + 1)) 2 1 e2 ~2 β= 2 2 3 2me c 4πǫ0 (na0 ) l(1 + 21 )(l + 1) ∂ 1 ∂2 1 ∂ 2 sin θ + L =− sin θ ∂θ ∂θ sin2 θ ∂ 2 φ
Es-o = β hS · Li =
2.1
(2.5)
−1
(2.6) (2.7) (2.8)
Angular-momentum eigenfunctions
From the table of spherical harmonics and Eq (2.2), (2.5), we have • h l1 m| l2 ni ≡ 0 if m 6= n, therefore h 11| 00i = 0 And also h 10| 00i = (constant) ×
Z
1
cos θ d(cos θ) = 0 −1
• For l = 1, 2, we only need to show that h 10| 20i = h 11| 21i = 0. Through inspections, the integrands as functions of cos θ are both odd, hence the integrals vanish.
2.2
Angular-momentum eigenfunctions
• According to Eq (2.1), l−1
Yl,l−1 = (−1)
s
2l + 1 i(l−1)φ l−1 e Pl (cos θ) 4π(2l − 1) 6
• It is convenient to write, h l, l − 1| l − 1, l − 1i = hl, l| (Lˆ− )† |l − 1, l − 1i = hl, l| Lˆ+ |l − 1, l − 1i =0
2.3
Radial wavefunctions
With n = 2, l = 1 the integral reads, Z ∞ 1 1 2 R (r)r 2 dr = 3 r r 3 2,1 0 Z ∞ Z 5 2 2 2 −r/a0 dr √ r e = r 2a0 3 0 = 1/(24a30 ) Invoking the rhs formula, namely
1 r3
1 = 1 l(l + 2 )(l + 1)
Z na0
2
(2.9)
yields the same result 1/(24a30 ).
2.4
Hydrogen
The probability is given by Z rb Z r 2 dr |ψ(r)|2 = 0
Z
rb 0
(r/a0 )2 d(r/a0 ) −2r/a0 e π r
ǫ:= ab
y 2 dy −2y e π 0 ≈ ǫ3 e−2ǫ /π ∼ (rb /a0 )3
=
0
The electronic charge density of this region is ρe ≈ e|ψ(rb /2)|2 =
2.5
e(1 − rb /a0 ) a30 π
Hydrogen: isotope shift, fine structure and Lamb shift
The mass ratio of electron to proton is ∼ 5 × 10−4 and isotope shift is of the same order, namely, if λ = 600 nm, order of isotope shift will be δ¯ νisotope ∼ (5 × 10−4 ) · (5 × 105 GHz) = 250 GHz 7
Relativistic effect is of the order of α2 ≈ 5 × 10−5 , then the wave number difference reads δ¯ νfs ∼ (5 × 10−5 ) · (5 × 105 GHz) = 25 GHz Lamb shift is 1/10 of the fine structure shift, that is δ¯ νLamb ∼ 2.5 GHz A Fabry-Perot ´etalon of finesse F = 100, width d = 1 cm is supposed to resolve 1 δ¯ ν= = 0.5 × 3 × 108 ∼ 0.1 GHz (2.10) n(2d)F but Doppler effect attenuates the resolution to ∼ 0.7 GHz ∼ 1 GHz. Therefore, isotope shift, fine structure can be resolved but Lamb shift approaches the limit of the apparatus and hence cannot be accurately observed.
2.6
Transitions
From Eq (1.12), we have Ultraviolet
100 nm
Infrared 1000 nm
2.7
0.45 ns 450 ns
Selection rules
Following similar arguments as in Problem 2.1 and notice there is an additional cos θ in the integrand.
2.8
Spin-orbit interaction
Calculations based on Eq (2.6) give Ej =
β l 2
β Ej ′ = − (l + 1) 2
and the mean of the two is ¯ = (2j + 1)Ej + (2j ′ + 1)Ej ′ = β[(l + 1)l − l(l + 1)] = 0 E
2.9
Selection rule for the magnetic quantum number
The integral is readily obtained in text of section 2.2.1, and the result follows the same arguments of Problem 2.1.
8
2.10
Transitions
(a) The wavefunction takes the form Ψ(t, r, θ) = Aψ1 eiE1 t/~ + Bψ2 eiE2 t/~ with A ≫ B, ψ1 = R1,0 Y00 , ψ2 = R2,1 Y1,0 and |Ψ|2 ≈ A2 |ψ1 |2 + 2|AB||ψ1∗ ψ2 | cos(ω12 t) The second term can be written as f (r)r cos θ cos(ω12 t) = f (r)r · zˆ cos(ω12 t) = f (r) z cos(ω12 t) The sketch of the orbital is as follows (Figure 1)
time
Figure 1: Contour of electron density, or the orbital of wavefunction Ψ(t, r, θ) = Aψ1 eiE1 t/~ + Bψ2 eiE2 t/~ with A ≫ B, ψ1 = R1,0 Y00 , ψ2 = R2,1 Y1,0 during one period of oscillation. ˆ |Ψi = ±1 |Ψi, namely the state exhibits parity, then (b) If Π ˆ |Ψi = − hri hri = hΨ| Πˆ† r Π hence hri = 0. But the Hamiltonian of hydrogen atom commutes with ˆ then the eigenstates are of specific parities. The only possibility for a Π, none-vanishing hri is to require ψ1 and ψ2 to possess opposite parities. (c)
Now set a0 = 1, the radial integral yields 5 Z ∞ 3 dr 2 1 √ r 4 e− 2 r = √ 6 6 3 0
The angular integral is exactly 1. The total electric dipole moment is 5 2 1 ea0 cos ωt zˆ −eD = − √ 3 6 where a0 is put back through dimension analysis. 9
(d) The density distribution should be more or less similar to Figure 1 for a constant φ but it becomes apparent if one writes c = ωt − φ or φ = ωt − c that a flock of charge is circulating about the z-axis. (e) The case of (a) is akin to the vertical motion of the electron while (d) corresponds to right-handed circular motion. The motions are characterized by the ml but the role of ground state 1s is crucial as pointed out in (b). The 1s state is an exponential decay which binds the electron to the region around nucleus and hence it is the quantum analogue of the classical restoring force −ω 2 r.
2.11
Angular eigenfunctions: Yll
(a) Raising operator is given as L+ = eiφ (∂θ + i cot θ∂φ )
(2.11)
Then we write (∂θ + i cot θ∂φ ) (Θeimφ ) = 0 it is equivalent to
Θ′ cos θ =m Θ sin θ
(b) The solution is Θ = sinm θ. Applying L2 (see (2.8)) yields L2 (Θeimφ ) = m(m + 1)Θeimφ
2.12
Parity and selection rules
If l1 − l2 is even, then Iang = (−1)l1 −l2 +1 Iang = −Iang which implies the integral vanishes. To have non-vanishing angular integrals, l1 − l2 must be odd. But the parity of spherical harmonics is ΠYlm = (−1)l Ylm
(2.12)
Therefore, the initial and final states must exhibit different parities.
2.13
Selection rules in hydrogen
The wavelengths corresponds to energy 0.306, 0.663, 1.890 eV and they are transitions from 5 to 4, 4 to 3 and 3 to 2 respectively. 10
3 3.1
Helium Estimate the binding energy of helium
(a) The total Hamiltonian is the sum of two individual and one interaction Hamiltonians, ˆ =H ˆ1 + H ˆ2 + H ˆ int H 2 2 ˆ 2i − Z e˜ ˆi = − ~ ∇ H 2m ri 2 e˜ ˆ int = H |r 1 − r 2 |
(3.1) (3.2) (3.3)
where e˜2 = e2 /4πǫ0 . (b) The energy E(r) =
Z e˜2 ~2 − 2mr 2 r
assumes minimum at rm =
~2 Zm˜ e2
and it is E(rm ) = − (c)
Z 2 m˜ e4 2~2
The repulsive energy, namely electron-electron interaction, is Eint =
e˜2 e˜2 Zm˜ e4 ≈ = r12 rm ~2
The ionization energy of one electron, according to the estimated energies, is Eion = 0 But experiment gives Eion = 24 eV, then the average distance between the two electrons should be greater than rm . (d) For Si12+ , Eion = (142 − 14 × 2) × 13.6 = 2285 eV, excluding the repulsion we have E = 2666 eV. Comparing with results of helium, repulsion gets irrelevant for larger Z.
3.2
Direct and exchange integrals for an arbitrary system
(a) The direct and exchange integrals are respectively, Z e˜2 J = d3 x1 d3 x2 |uα (r1 )|2 |uβ (r2 )|2 r12 Z 2 e ˜ K = d3 x1 d3 x2 u∗α (r1 )u∗β (r2 ) uα (r1 )uβ (r2 ) r12 11
(3.4) (3.5)
Yet, |Ai = (2)−1/2 (|αβi − |βαi), 1 (hαβ| − hβα|)Hˆ ′ (|αβi − |βαi) 2 hαβH ′ αβi − hαβH ′ βαi + α ↔ β = 2
ˆ ′ |Ai = hA| H
Note that
(3.6) (3.7)
αβH ′ αβ = βαH ′ βα
∗ αβH ′ βα = βαH ′ αβ
Therefore, it holds for real-valued u(r) that
ˆ ′ |Ai = αβH ′ αβ − αβH ′ βα = J − K hA| H
(b) The symmetric wavefunction is construct as
|Si = (2)−1/2 (|αβi + |βαi) The inner product reads h A| Si =
(1 − 1 + h αβ| βαi − c.c) = Im h αβ| βαi 2
Again, for real u(r), h A| Si are orthogonal. ˆ ′ is invariant under the interchange of particle labels, let Σ (c) Since H denote such an operation, then ˆ ′ Σ |Si = − hA| Hˆ ′ |Si hA| Hˆ ′ |Si = hA| Σ† H ˆ ′ |Si = 0. Hence hA| H
3.3
Exchange integrals for a delta-function interaction
(a) The Hamiltonian is simply the kinetic energy, H=−
~2 2 ∂ 2m x
For the fact that u′′0 = −(π/l)2 u0
u′′1 = −(2π/l)2 u1 ,
then the energies are E0 =
~2 π 2 2ml2
E1 = 12
2~2 π 2 ml2
(b) The direct integral is 2 Z 2 πx2 πx1 J= )aδx1 −x2 sin2 ( ) dx1 dx2 sin2 ( l l l 2 Z l 2 sin4 (πx/l)dx =a l 0 Z 4 π 4 ′ ′ 3a =a sin x dx = πl 0 2l The exchange integral is the same thing, K = J. Then the energy shift will be 3a/l. There is only one state, the symmetric one. (c)
All the eigenstates takes the form, r nπx 2 sin un (x) = l l
and trivially Jmn = Kmn since the delta potential identifies the two coordinates. Hence, the antisymmetric state gives no rise to energy shift. Actually, the antisymmetric part does not exist at all when interaction is considered, because delta interaction rules out the possibility of ‘no-touch’, otherwise there is no interaction, and hence no energy shift! (d) See Figure 2.
1 0.6
0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8
0.2 -0.2 -0.6 -1 1 0.8 0.6
1 0.8
0.4
0.6
x2
0.4
x1
0.2 0.2 00
Figure 2: The horizontal axes are the coordinates x1 and x2 , and the vertical axis marks the values of function ψ = u1 (x1 )u2 (x2 ).
(e) Possible total spins are 0 and 1. The symmetric spatial functions corresponds to 0-spin, and the antisymmetric spatial functions should have spin-1 or spin-0 but they do not exist.
13
(f ) The wavefunctions are independent of particle masses, therefore all the mathematics are invariant once the states are given. The energy levels are still E+ = E1 + E2 + 2J E− = E1 + E2 but the difference is that the levels are not related to the exchange symmetries of the particles.
3.4
A helium-like system with non-identical particles
The exoticon-exoticon system (identical fermions) is indifferent with the electron-electron system except for richer spin configurations for the two symmetries. Yet no restriction upon symmetries of particle exchange is placed on exoticon-electron system, therefore, the spatial orbitals can be freely occupied by the two fermions, and for the fact that spin is not included in the Hamiltonian, hence the energies levels are left unaltered.
3.5
Integrals in helium
Set a0 = 2Z, the integral in the curly brackets is −r2 e−2r2 /2 − e−2r2 /4 + 1/4 and J=
3.6
e˜2 5 Z = 34 eV 2a0 4
Calculation of integrals for 1s2p configuration 6
R R10 20 R 21
5
4
Rnl
3
2
1
0
-10
2
4
r / a0
6
8
10
Figure 3: The plot of R10 , R20 and R21 with Z = 2. The integral J1s2p = −
e˜2 × 0.00208 = 0.0283 eV 2a0 14
3.7
Expansion of 1/r12
Expansion in terms of spherical harmonics is the following ∞ k 1 1 X r1 k 4π X ∗ = Ykq (θ1 , φ1 )Ykq (θ2 , φ2 ) r12 r2 r2 2k + 1 k=0
(3.8)
q=−k
(a) Setting k = 0, 1, we have 1 1 ≈ {4π/4π r12 r2 r1 4π 3 + (cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 + φ2 )) r2 3 4π 1 r1 = 1 + cos θ12 r2 r2 (b) Mathematically, the expectation values of 1/r12 are always sandwiched by ‘bra-ket’ where the phase factors cancels; this is seen in the expression for K1s nl . Physically, the quantum number m is responsible for magnetic interactions but Coulomb repulsions have no interests in that. (c) Due to the orthogonality relations of spherical harmonics, the terms in 1/r12 are eliminated excluding the one with quantum numbers lm. For l = 1, this corresponds to the second term in the expansion as shown in part (a). (d) This follows the arguments in part (c), the given l samples out the order k in Eq (3.8).
4
The alkalis
4.1
Configuration of the electrons in francium Fr = [Rn]7s1 Rn = [Xe]5d10 5f 14 6s2 6p6 6d10 Xe = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f 14 5s2 5p6
4.2
Finding the series limit for sodium
Energy levels of sodium obey the rule E=−
Ry 13.6 eV =− (n∗ )2 (n − δ)2 15
(4.1)
Then ∆E ∝ E 3/2
(∆E)2 ∝ E 3
The topmost level must have ∆E = 0, and that energy corresponds ionization energy which was found to be 5.1 eV. The effective principal number p ∗ n = 13.6/5.1 = 1.63.
4.3
Quantum defects of sodium
From formula Eq (4.1), the quantum defects are 3s 1.37
4s 1.34
5s 1.33
6s 1.35
the average quantum defect is 1.35 ± .02. Assume that the quantum defect for 8s is still 1.35, the binding energy will be ENa 8s = −13.6/(8 − 1.35)2 = 0.31 eV while for hydrogen EH 8s = 0.21 eV.
4.4
Quantum defect
Quantum defect of Rb 5s is 3.19 with which that of 7s is approximated. Therefore the energy difference reads 1 1 ∆E = 13.6 − = 3.22 eV (5 − 3.19)2 (7 − 3.19)2 Then the wavelength of the two identical photons are λ=
4.5
hc = 771 nm ∆E/2
Application of quantum defects to helium and heliumlike ions
Throught a direct calculation, the wavelength from 1s3d to 1s2p is obtained as 625 nm, compared with 656 nm of hydrogen Balmer-α. Quantum defects can be evaluated by Eq (4.1), they are listed in Table 2. It is readily seen that δs > δp > δd . Table 2: Quantum defects of helium 1s2s 1s2p 1s3s 1s3p 1s3d 0.2356 0.0275 0.2265 0.0287 0.0026 To estimate binding energy of 1s4l states, assume the quantum defects for l = 1, 2, 3, 4 as 0.23, 0.28, 0.0026, 0.00. Then the binding energies read 0.96, 0.98, 0.85, 0.85 eV 16
The second ionization energy of Li+ is IE2 = −(E ′ − E1s4f ) = 72.24 +
Z 2 Ry = 75.64 eV 42
where Z = 2.
4.6
Quantum defects and fine structure of potassium
The wave numbers can be put into three groups (3 row vectors), and take average and difference within each group. The data set becomes ν = (hνi , δν). The energies are calculated through E = hc/λ, and the data set for energies reads E = (hEi , δE). Given that the ionization energy of potassium IE = 4.34 eV, the energy levels are E − IE from with we can find out the effective principal quantum numbers and quantum defects. They are n∗ = 2.23, 3.26, 4.28 and δl = 1.77, 1.73, 1.72; the pattern confirms that they corresponds to 4p, 5p, 6p. The fine structure splitting δE/E divided by α2 are, for 4p, 5p, 6p, 49.18, 33.45, 26.38. According to Land´e formula, ∆Efs =
Zi2 Zo2 α2 Ry (n∗ )3 l(l + 1)
(4.2)
the ratio of ∆E/(n∗ )3 should be 1:1:1 and it turns out to be 1.00 : 0.99 : 1.03. The transitions of 7p should produce spectral lines of wavelengths 321.84 nm, 321.93 nm.
4.7
The Z-scaling of fine structure
From Eq (4.2), fine structure splitting of ions scales as Z 4 , then for it is 1.3 × 10−5 × 114 = 0.19 eV; for neutral ions it scales as Z 2 , namely, for sodium atom it should be 0.0016 eV.
Na+10 ,
4.8
Relative intensities of fine-structure components
(a) In this problem J = 1/2, 3/2, 5/2, hence the weight ratio should be 2 : 4 : 6 = 5 : 10 : 15 = 5 : (1 + 9) : (1 + 9 + 5). (b) Possible electrical dipole transitions are D5/2 −F5/2 , D5/2 −D7/2 , D3/2 − F5/2 with intensities 1 : 20 : 14. Hint: Plot out energy levels and then count and take ratios.
4.9
Spherical symmetry of a full sub-shell
Show that sum reads
Pl
m=−l
|Ylm (θ, φ)|2 is spherically symmetric. For l = 1, the s1 =
3 3 (cos2 θ + sin2 θ) = 4π 4π 17
Associated Legendre polynomials are defined as follows m d Plm (x) = (−1)m (1 − x2 )m/2 Pl (x) dx
(4.3)
and they are related to spherical harmonics in Eq (2.1).
5
The LS-coupling scheme
5.1
Description of the LS-coupling scheme
Central-field approximation is achieved by imposing spherical symmetry on electron distributions where electrons are seated in several spherically symmetric layers. The central field Hamiltonian is the sum of individual electron Hamiltonian and hence the Shr¨odinger equations are decoupled. The arrangements of pseudo-independent electrons constitute the electron configurations. (a) The residual electrostatic interactions couples the electrons and give rise to splitting of energy levels. (b) Magnetic spin-orbit interactions include another degree of freedom, the spin, which ‘rotates’ the original eigenstates resulting in further splittings.
5.2
Fine structure in the LS-coupling scheme
The spin-orbit Hamiltonian reads hl1 · Li hs2 · Si hl2 · Li hs1 · Si S· + β2 S· := βLS S · L S(S + 1) L(L + 1) S(S + 1) L(L + 1) (5.1) 3 For 3s4p P configuration, S = 1 = L, therefore the relation βLS = β4p /2 holds if 2 1X β4p = βi hsi · Si hli · Li 2 Hs−o = β1
i=1
but consider the symmetry of electrons, the right hand side is just X βi sik lik = β4p i
5.3
The LS-coupling scheme and the interval rule in calcium
The ground configuration of calcium: 1s2 2s2 2p6 3s2 3p6 4s2 . The triple lines comes from 3 P term with J = 0, 1, 2 with and interval ratio ∼ 2. The first three of muiltiplet of six lines come term 3 DJ=1,2,3 with an interval ratio ∼ 1.5; the rest must be 3 P, 3 D and 3 F from selection rules.
18
5.4
The LS-coupling scheme in zinc
The zero comes from 1 S0 , then it leaps to triplet 4s4p3 P0,1,2 , next it jumps to singlet 4s4p1 P1 and 4s5s 3 S1 , 1 S0 .
5.5
The LS-coupling scheme
The interval rule accurately indicates that the four levels of Mg are 3 P0,1,2 and 1 P1 of 3s3p. Heavier Fe14+ shows a weaker interval pattern (2.4 rather than 2) due to intercombination and therefore ∆S 6= 0 transitions occurs in Fe14+ but not Mg.
5.6
LS-coupling for configurations with equivalent electrons
(a) For np2 configuration (2 out of 6 states, 15-fold degenerate), ML = ml1 + ml2 and MS = ms1 + ms2 , then ML = ±2, ±1, 0 and MS = ±1, 0. Considering Pauli exclusion, the 3 D and 3 S can be directly eliminated preserving 1 D (5-fold) and 1 S (1-fold). Then what’s left is 15 − 5 − 1 = 9 and 3 P (9-fold) fits. (b) The first 3 lines are the spin-orbit splittings of 3 P term, the latter two are 1 D and 1 S. The weak emissions lines near 1 D indicates deviations from LS-coupling scheme. (c) In order to make MS = 2 and ML = 2 for six d-electrons, ms = 1 1 1 1 1 ( 2 , 2 , 2 , 2 , 2 , − 12 ) and ml = (2, {0}5 ) where curly bracket means ‘all combinations.’ To meet Pauli exclusion principle and maximize MS , the longest sequence of aligned ms is up to 5 since ld = 2, namely, five electrons are equally spread out, therefore ms takes the form above and ML = ml (ms = 1 2 ) ≤ ld = 2.
5.7
Transition from LS- to jj-coupling
For configuration 3p4s, Hre > Hs-o due to a higher correlation energy (exchange integral) from inner core, while 3p7s is just the opposite, the electrons seems more independent. Therefore, LS-coupling is proper for 3p4s and jjcoupling for 3p7s. The former gives two terms 3 P and 1 P while the latter gives four levels: (3/2 = (1+1/2), 1/2)J=2 , (1/2 = (1−1/2), 1/2)J=1 , (1/2 = (1 − 1/2), −1/2)J=0 , (3/2 = (1 + 1/2), −1/2)J=1 . It is shown in Figure 5.10 in the text.
5.8
Angular-momentum coupling schemes
Take the 0-2 and 1-1 difference of np(n + 1)s for n = 3, 4, 5, then take their ratios, I find ∆02 /∆11 = 0.23, 0.70, 0.89. As n grows, the LS-coupling transitions to jj-coupling, namely ∆02 /∆11 → 1. The g-factor gJ = 1.06 should belong to 1 P (spin-0), the deviation from 1 is due to the mixing with other 3 P wavefunctions. 19
5.9
Selection rules in the LS-coupling scheme
(a) No, ∆l = 2; (b) no, ∆J = 2; (c) yes; (d) no, ∆l = 0; (e) no, ∆J = 2. 4d9 5s5p is the candidate that mixes with 4d10 5p.
5.10
The anomalous Zeeman effect
Electric dipole transition rules require ∆MJ = 0, ±1. The energy is given by EZM = gJ µB BMJ (5.2) The transitions 3 S1 (gJ = 2)-3 P2 (gJ = 23 ) take place at MJ : [1, 0, −1] → [2, 1, 0, −1, −2], then (gJ MJ ) : [2, 0, −2] → [3, 23 , 0, − 32 , −3]. Therefore, ∆(gJ MJ ) = [(1, 32 , 2); (− 21 , 0, 12 ); (−2, − 23 , −1)] The spacing is µB B/2 = 14 × 1/2 = 7 GHz with B = 1 T
5.11
The anomalous Zeeman effect
Formula for the g-factor: gJ =
3 S(S + 1) − L(L + 1) + 2 2J(J + 1)
(5.3)
then gJ [ 3 S1 ] = 2 and gJ [ 3 P1 ] = 23 . Compared with the previous problem (or Figure 5.13), since ∆J = 0, ∆MJ 6= 0, there are only six lines left.
5.12
The anomalous Zeeman effect
With previous experience and the fact J = 0 states have no Zeeman splitting, possible transitions are: 3
P1 → 3 D2 (1),
3
P2 → 3 D1 (2)
and the g-factors are 3
P1 = 3/2
3
3
D1 = 1/2
3
P2 = 3/2 D2 = 7/6
The quantities (gJ MJ )1 : [3/2, 0, −3/2] → [7/3, 7/6, 0, −7/6, −7/3] and (gJ MJ )2 : [3, 3/2, 0, −3/2, −3] → [1/2, 0, −1/2] Therefore, 6∆(gJ MJ )1 = [+9, +7, +5; +2, 0, −2; −5, −7, −9] fits the data.
20
5.13
The anomalous Zeeman effect in alkalis
(a) gJ [( 2 S1/2 , 2 P1/2 , 2 P3/2 )] = (2, 2/3, 4/3) (b) The g-factors are gJ [3s 2 S1/2 ] = 2 gJ [3p 2 P3/2 ] = 4/3 It is a set of ‘2 to 4’ transitions, then there are six lines. The intervals are 4/3×[3/2, 1/2, −1/2, −3/2]−2×[1/2, −1/2] = [(5/3, 1)(1/3, −1/3)(−1, −5/3)] and they are equally spaced by 2/3 µB B = 2/3 × 14 × 1 = 9.33 GHz (c) There are only two lines of same intensity, no ∆MJ = 0 transition. (d) The energy of fine structure is 517.96 GHz this corresponds to a huge magnetic field 517.96/9.33 = 55.5 T.
5.14
The Paschen-Back effect
Spin, in this case, is a spectator variable, then J = L, hence gJ = 3/2 − 1/2 = 1, same as in the normal Zeeman effect.
6
Hyperfine structure and isotope shift
6.1
The magnetic field in fine and hyperfine structure
The magnetic field is given by 2 B e = − µ0 gs µB |ψns (0)|2 s 3 where |ψns (0)|2 =
(6.1)
Z3 πa30 n3
Setting Z = 1, s = 1/2 and n = 1, 2 for 1s and 2s, we have magnetic field flux density at center as 16.7 T and 2.1 T respectively. The magnetic field felt by an orbiting electron is B = βl where β is the spin-orbit interaction constant given by Eq (2.7). Then the field that 2p-electron in hydrogen experiences is 0.2 T.
6.2
Hyperfine structure of lithium
Hyperfine and fine structure splittings are both proportional to the magnetic moment (spin) and in turn, proportional to the mass. Therefore, hyperfine structure is of order me /Mp smaller than fine structure. Let P = max{I, J} and Q = min{I, J}, then F = {P + Q, P + Q − 1, . . . , P − Q + 1, P − Q} and hence there are 2Q + 1 values of F . 21
Since 2s has no fine structure, the fine structure totally comes from 2p 2 P3/2,1/2 with maximum J = 3/2 on the top. For 6 Li (at J = 3/2), 2I + 1 = 3, then I6 = 1 and for 7 Li, 2I + 1 = 4, I7 = 3/2, and notice both are no larger than 3/2. Interval Rule With computer, I find the splittings for I = 1 is 3 : −2 : −5, the interval ratio is then 5 : 3; for I = 3/2, the splittings are 4.5 : −1.5 : −5.5 : −7.5 with interval ratio 3 : 2 : 1. The data confirms our prediction of nuclear spins given the hyperfine constants for e to g negative. The hyperfine shift of J = 1/2 level are 1 : −2 and 1.5 : −2.5, then the gap ratio will be 3 : 4. From the 2s hyperfine structures, we can find out the gI ratio of two isotopes, it is (228.2/803.5)(4/3) = 1/2.641. Then we have X = 4 × (26.1/3) × 2.641 = 91.899 = 91.9 MHz
6.3
Hyperfine structure of light elements
The hyperfine structure can be estimated as EHFS ∼
Zi Zo2 me 2 α Ry (n∗ )3 Mp
(6.2)
For ground state hydrogen, the hyperfine structure should be around 95 MHz and for ground state lithium (Zi = 3, Zo = 1, n∗ = 1.59), 71 MHz. Both structures are badly underestimated.
6.4
Ratio of hyperfine splittings
For 1s state, J = 1/2, put in notation {element : I; F } we have relevant quantum numbers for H, D and 3 He+ : {H : 1/2; 1, 0}
{D : 1; 3/2, 1/2}
{3 He+ : 1/2; 1, 0}
Given A ∝ gI µN Z 3 , gI ∝ µI /I and interval rule, we have EH /ED =
3/2 × 2.79 AH = = 4.3 3/2 AD 3/2 × 0.857
EH /E 3 He+ =
6.5
2.79 AH = −0.16 = A 3 He+ −2.13 × 8
Interval for hyperfine structure
(a) The interaction can be expressed in terms of F, I, J, EF = A hI · J i =
A {F (F + 1) − I(I + 1) − J(J + 1)} 2 22
(6.3)
Therefore, EF − EF −1 = A2 {F (F + 1) − F (F − 1)} = AF , the interval rule. (b) Apply the same method as the text demonstrated, F = 4.9, 6.1, 6.9, 8.1, respectively for c, d, e, f. The nuclear spin is I = Fmax −J = 8−11/2 = 5/2. (c) From the peaks, it can be inferred that nuclear spin is 5/2 for both isotopes, then A( 8 S7/2 ; 153) = 20 × (4.86 − 2.35)/(6.42 − 0.77) = 8.9 MHz.
6.6
Interval for hyperfine structure
From the argument in (6.2), I postulate that 2I +1 = 6 and I = 5/2. The splittings make the ratio 3.1 : 2.6 : 2.0 : 1.5 : 1.0 consistent with prediction 3.0 : 2.5 : 2.0 : 1.5 : 1.0.
6.7
Hyperfine structure
From the data: (a) 70/5 = 42/3 = 14, (b) 70/42 = 5/3, I conclude that 70 and 42 are two levels of 39 K and 5 and 3 the corresponding two levels of 40 K. Since the intensities depend on degeneracies 2F + 1, then the two levels are F = 1 and F = 2. Then nuclear spin is just I = 2 − 1/2 = 3/2, consistent with the number of splittings (I > J). And the magnetic moment (µI = gI I, I the same) ratio (39 over 40) reads 1.6/(1.9 − 1.0) = 1.8.
6.8
Zeeman effect on HFS at all field strengths
(a) As B goes up, the atom moves from F - to J-scheme, and from the numbers of splittings, we have 2J + 1 = 4, J = 3/2 and 2F1 + 1 = 5, 2F2 + 1 = 3, F = 2, 1. Hence I = 2 − 3/2 = 1/2. (b) MJ is the good quantum number. (c) The weak field Zeeman energy is EZM, weak = gF µB BMF
(6.4)
Same separations is equivalent to same gF which is gF =
F (F + 1) + J(J + 1) − I(I + 1) hJ · F i gJ = gJ F (F + 1) 2F (F + 1)
(6.5)
J = 3/2, I = 1/2, F = 2, 1, from calculation the gF are different. (d) The strong field Zeeman energy is EZM, strong = gJ µB BMJ + AMI MJ
(6.6)
If µB B ≫ A, and J is same for both hyperfine states, then Zeeman splittings are the same. (e) The cross-over can be defined at µB B = A = 3.4 GHz, or B = 3.4/14 = 0.24 T.
23
6.9
Isotope shift
The mass effect shifts the energy by ∆˜ νMass ≈
me δA ν˜∞ Mp A′ A′′
(6.7)
while volume effect gives ∆˜ νVol
2 rN δA Z 2 ≈ R∞ a20 A (n∗ )3
(6.8)
and radius of nucleus is rN ≈ 1.2 × A1/3 fm
(6.9)
For rubidium isotopes A′ = 85 and A′′ = 87, then total isotope shifts, mass plus volume effects, read (a) .186 + .590 = .777 m−1 (b) .202 + .590 = .792 m−1 .
6.10
Volume shift
From Eq (6.8), the relative uncertainty that rN contributes is δ[∆˜ νVol ] = 2δ[rN ],
d[∆˜ νVol ] = 2% ∆˜ νVol
For hydrogen 1s configuration, the volume shift is about 5 × 10−9 eV ≈ 1 MHz. Therefore, d[∆EVol ]/ELamb ≈ 1% × 1/1057.8 ∼ 1 ppm.
6.11
Isotope shift
Assume Z = A/2, n∗ = 2, ν˜∞ = 1/500 nm−1 , then A11/3 =
6.12
me ν˜∞ R∞ /a30 × 32/(1.2 × 10−15 )2 Mp
A = 71
Specific mass shift
P Substitute the nucleus momentum for pN = − N i pi , ergy becomes N N X me X 2 1 1+ pi · pj T = pi + 2 2me MN i
then kinetic en-
specific mass effect
24
6.13
Muonic atom
The radius is easily found to be r=
n2 ~2 = a0 /2277 Zmµ e˜2
since n = 1, mµ = 207me and Z = 11. The ‘muonic Bohr radius’ is way much smaller than electronic Bohr radius. The binding energy is En = −
Z 2 e˜4 mµ = −112 × 207 Ry/n2 2n2 ~2
If n = 1, E1 = −25047 Ry. Volume effect must be extremely evident since the size of orbit is comparable to nucleus scale. Actually, it is about 4% of the transition energy.
7
The interaction of atoms with radiation
7.1
Averaging over spatial orientations of the atom
(a) We need to show that | hxi |2 = | hri |2 /3. Proof. Z
2
dΩ (cos φ sin θ) =
Z
1
−1
d(cos θ) (1 − cos θ)
= y− 4π = 3
2
1 y3 3
−1
1 = 3
×
Z
Z
2π
0
dφ
1 + cos(2φ) 2
2π 2
dΩ.
(b) Since there is no specific preference of coordinate system, the result in (a) is valid for all three axes and hence any direction in space.
7.2
Rabi oscillations
(a) The proof is straightforward for equivalence of eqns 7.25 and 7.26. The equation 7.26 allows for a family of solutions with arbitrary constant phases, without loss of generality, set |Ω| Wt c2 = sin = const × (eiW t/2 − e−iW t/2 ) W 2 and plug it into the equation 7.26 2 Ω c¨2 + i(ω − ω0 )c˙2 + c2 = 0 2 25
(7.1)
The first exponential gives −(W/2)2 − (ω − ω0 )W/2 + |Ω/2|2 = 0 the second exponential (with minus sign) gives (W/2)2 + (ω − ω0 )W/2 − |Ω/2|2 = 0 Both cases are consistent, if ω ≈ ω0 , with definition of W , W 2 = Ω2 + (ω − ω0 )2
(7.2)
(b) The plot. -2
0
2
4
6
8
10
12
14
1.2
1.2 F1 F2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-0.2
-0.2 -2
0
2
4
6
8
10
12
14
Figure 4: F1 for Ω and F2 for 3Ω
7.3
π- and π/2-pulses
(a) Try the ansatz c(t) = c exp(iλt/2), I find λ = Ω, then with initial conditions c(0) = [1, 0], c1 (t) = cos(Ωt/2)
c2 (2) = sin(Ωt/2)
(b) A general solution takes the form, Ωt Ωt |ψi = cos + φ |1i − i sin + φ |2i 2 2 Initially, |ψ(0)i = cos φ |1i − i sin φ |2i 26
(7.3)
Now set Ωt = π, then |ψ(tπ )i = − sin φ |1i − i cos φ |2i = −i{cos φ |2i − i sin φ |1i} (c) Set Ωt = 2π, |ψ(t2π )i = − |ψ(0)i. (d) Set φ = 0 and Ωt = π/2, √ φ(tπ/2 ) = (|1i − i |2i)/ 2
(e) First set φ = 0, Ωt = π/2, then the state becomes √ ′ 1 = {|1i − i |2i eiϕ }/ 2
with an orthogonal state √ ′ 2 = {−ie−iϕ |1i + |2i}/ 2.
Here comes a second π/2-pulse, the state evolves into 1 + eiϕ 1 − e−iϕ |1i − i |2i 2 2
Therefore, the probabilities in |1i and |2i√are sin2 (ϕ/2) and cos2 (ϕ/2). (f )√Since cos(φ + π/4) = (cos φ − sin φ)/ 2 and sin(φ + π/4) = (cos φ + sin φ)/ 2, then 1 1 −i ˆ U π2 = √ 2 −i 1 Similarly,
and the phase shift in |2i
ˆπ = U
ˆϕ = U
1 0 0 eiϕ
0 −i −i 0
Then the successive operator reads ˆϕ U ˆπ U ˆπ ˆ =U ˆπ U U 2 2 ϕ cos 2 = −eiϕ/2 − sin ϕ2
7.4
sin ϕ2 cos ϕ2
The steady-state excitation rate with radiative broadening
(a) The differential equation reads c˙2 = − Γ2 c2 , the solution reads, from inspection, c2 (t) = c2 (0)e−Γt/2 .
27
(b) Since
integration gives
d Γt/2 Ω∗ = −ic1 e−i(ω−ω0 +iΓ/2)t c2 e dt 2 c2 (eΓt/2 − 1) = −i
Ω∗ e−i(ω−ω0 +iΓ/2)t 2 Γ/2 − i(ω − ω0 )
where c1 = 1 in the weak excitation. In the limit Γt ≫ 1, |c2 |2 =
7.5
Ω2 /4 (ω − ω0 )2 + Γ2 /4
Saturation of absorption
(a) Transmission in a weak resonant field takes the simple form I(ω, z) = I(ω, 0)e−κ(ω)z = I(ω, 0)e−N σ(ω)z
(7.4)
For transmission 1/e, N σ = 1. The absorption cross-section reads σ(ω) =
g2 π 2 c2 A21 gH (ω) g1 ω02
(7.5)
where g1 , g2 are the degeneracies of two states and gH is the Lorentzian line shape function Γ 1 (7.6) gH (ω) = 2π (ω − ω0 )2 + Γ2 /4 For a simple two level system (Γ = A21 ) with g2 /g1 = 3 at resonance, σ(ω0 ) = 3 ×
2πc2 A21 3λ20 = 2π ω02 Γ
(7.7)
Sodium s-p transition has λ0 = 589 nm, invoking Eq (7.7), we have N = 6 × 1012 . (b) The saturation intensity is defined as Is (ω) =
~ωA21 2σ(ω)
(7.8)
and the fact that Isat = Is (ω0 ), Isat /Is = ω0 σ(ω0 )/ωσ(ω). The absorption coefficient is defined N σ(ω) κ(ω, I) = (7.9) 1 + I/Is (ω) then we can find κ(ω, Isat ). Notice τ = 1/Γ = 6 ns for sodium at λ = 589 nm.
28
7.6
The properties of some transitions in hydrogen
(b) The A21 for 3s, 3p and 3d are 6.3 × 106 s−1 , (17 + 2.2) × 107 s−1 and 6.5 × 107 s−1 , the lifetimes τ = 1/A21 are then 160, 5.5, 15 ns. (c) Shorter lifetime means larger spontaneous transition rate. Since 1s2p overlaps much more than 1s-3p, transition rate is higher for the former and hence shorter lifetime. (d) The Einstein coefficients satisfy the following relations A21 =
~ω 3 B21 π 2 c3
g1 B12 = g2 B21
(7.10)
and radial matrix element D12 is contained in B12 =
πe2 |D12 |2 3ǫ0 ~2
⇒
A21 =
g1 4α 3 ω |D12 |2 g2 3c2
(7.11)
Then the D12 ’s for transitions 2p-3s, 1s-3p, 2s-3p, 2p-3d and 1s-2p are 0.53, 0.52, 3.0, 3.8, 1.3, measured in a0 . (e) The saturation intensity is given by Isat =
π hcΓ π hc = 3 λ3 τ 3 λ3
(7.12)
For the case where spin and fine structures are ignored, Γ = A21 . Now calculate the saturation intensities for 2p-3s and 1s-3p. The A21 are (6.3 × 106 , 1.7 × 108 ) s−1 , the wavelengths are 1./R∞ (5/36, 8/9), therefore Isat = (0.46, 3.7 × 103 ) mW/cm2 .
7.7
The classical model of atomic absorption
The dynamical equation takes the form x ¨ + β x˙ + ω02 x =
F (t) cos ωt m
(7.13)
The steady state ansatz takes the general form x(t) = Aei(ωt+δ) and plug it into the dynamical equation (I have secretly replaced cos ωt by eiωt ), we have F (t) −iδ e −ω 2 + iβω + ω02 = mA (a) Take the real part of the ansatz, x = A cos(ωt + δ) = A(cos ωt cos δ − sin ωt sin δ)
: = u cos ωt − v sin ωt
29
(b) Notice that p
u2 + v 2 = A F0 /m e−iδ = 2 ω − ω 2 + iβω 0
=p
F0 /m
− ω02 )2 + (βω)2 −1/2 β2 F0 2 (ω − ω0 ) + ≈ 2mω 4 (ω 2
The peak is located at ω 2 = ω02 − β 2 /2. (c) The phase is given by tan δ = =
v u ω2
βω − ω02
The first line comes from (a) and the second line comes from the complex algebraic equation above. (d) Combine (b) and (c). (e) Notice that P ∝ |A|2 and simply invoke (b) we have, P ∝
7.8
1 (ω − ω0
)2
+ (β/2)2
Oscillator strength
(a) Integrate the cross-section in Eq (7.5) gives Z ∞
g2 π 2 c2 g1 4α σ(ω)dω = |D12 |2 ω 3 2 2 g1 ω0 g2 3c −∞ π 2 c2 4α |D12 |2 ω03 ≈ 2 ω0 3c2 = 2π 2 r0 cf12 where, as you may check, r0 = α~/me c = 2.8 × 10−15 . Note, I computed the complicated integral with computer algebra system and made approximations by hand. (b) The cross-section can be calculated from σ(ω) =
P (ω) 8π/c e4 ω 4 = 2 2 2 3 (c/8π)E0 3m3 c (ω − ω02 )2 + β 2 ω 2
30
(c) For an harmonic oscillator E = K + U = 2K, if we assume that oscillation is much faster than the damping, then this formula holds and we have
E = m x˙ 2 = mx20 e−βt ω ′4 cos2 (ω ′ t) = mx20 e−βt ω ′4 /2
Then, according to the classical damping formula (1.23), the damping rate is E˙ e2 ω 2 1 =− =β= τ E 6πǫ0 me c3 (d) Combine the result in (b) and (c). (e) From Eq (7.11), we have |D12 |2 =
g2 3λ3 A21 g1 4αc(2π)3
(7.14)
and for sodium 3s-3p transition, A21 = 2π × 107 s−1 , λ = 589 nm, therefore |D21 |2 = 5.3 × 10−20 C2 m2 . Absorption oscillator strength is, by definition, 2 f12 = 2me ω0 D12 /(3~) = 0.980
(7.15)
(f ) For hydrogen 1s-2p and 1s-3p the oscillator strengths follow from the formula 3 mc2 A21 f12 = 2 ~αω 2 and they are 0.418, 0.0801.
7.9
Doppler broadening
In the rest frame and moving frame the frequency differs by δ = ω −ω0 = kv. And the sinc2 behaves like a delta, then v = (ω0 − ω)/k in f (v).
7.10
An example of the use of Fourier transforms
I recommend first complexify the expression, then apply Fourier transform and finally take the real part. It is straightforward if you look it up in the table and the final result will be a Lorentzian with peak at ω.
7.11
The balance between absorption and spontaneous emission
It comes from the Einstein equation N˙ 2 = (N1 B12 − N2 B21 )ρ(ω12 ) − N2 A21
(7.16)
at equilibrium and if we neglect the contribution from spontaneous emission, the absorption is κI = (N1 B12 − N2 B21 )ρ(ω12 )~ω = N2 A21 ~ω 31
Note that N2 = N ρ22 and substitute eqn (7.69) for ρ22 , then we have eqn (7.87) invoking the definition I Isat
7.12
=
2Ω2 Γ2
(7.17)
The d.c. Stark effect
(a) The perturbation is similar to “intermediate field strength” in the discussion of Zeeman effect. The eigenvalues are λ1,2 = ±
p
ǫ V2 ) V 2 + ǫ2 /4 ≈ ±( + 2 ǫ
(b) See part (a). (c) For 3s-3p transition, ǫ = hc/λ = 2.11 eVThe displacement will be of the order a0 , then the electrostatic energy will be of the order V = eE0 a20 /2 ∼ 10−18 eV. Then ∆E = V 2 /ǫ ∼ 10−37 eV.
32
