solutions to the exercies of -...
TRANSCRIPT
Solutions to the Exercies ofC. J. Foot’s Atomic Physics
Chenchao ZhaoDepartment of Physics, Beijing Normal University, Beijing, China
(Dated: June 28, 2011)
1 Early atomic physics
Key formulas:
1
λ= R
(1
n2− 1
n′2
)
(1.1)
RH = RmH
me +mH(1.2)
a0 =~2
(Ze2/4πǫ0)me= 0.529 × 10−10 m (1.3)
E = −Ze2/4πǫ02a0
1
n2∝ Z2 (1.4)
∆E =α2
n2E (1.5)
α =e2/4πǫ0
~c= 1/137 (1.6)
f
c= R∞((Z − σK)2 − (Z − σL)
2) (1.7)
ρ(ω) =~ω3
π2c31
exp ~ω/kT − 1(1.8)
N2
g2=N1
g1exp
(
−~ω
kT
)
(1.9)
P =e2x20ω
4
12πǫ0c3(1.10)
ΩL =eB
2me(1.11)
τ =6πǫ0mec
3
e2ω2(1.12)
1
1.1 Isotope shift
To find wavelengths of Balmer-α transitions, we set n = 2 and n′ = 3 asin Eq (??) Then λ = 36
5R , hence
λH − λD =36
5(1/RH − 1/RD)
=36me
5R∞
(1
mH− 1
mD
)
≈ 18me
5mHR∞= 0.18 nm
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, ne-glecting the isotope shifts, the energy levels in agreement are those with1/n2 = 4/m2, namely m = 2n.
Those wavelengths should have the ratio
λHλHe
=RHe
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
∆λ=
E
∆E=
4
α2= 75076 (1.13)
for the fact that λ = ch/E and
dλ = −ch dE/E2
= −(ch/E) (dE/E)
= −λ (dE/E)
This corresponds to a grating of ∼ 105 grooves.
2
1.4 X-rays
Eq (1.7) reduces to f = 3cR∞Z2/4 when Z ≫ σ. Therefore,
√f ∝ Z
1.5 X-rays
Since E = hf = 3/4(hcR∞)Z2 = 13.6 eV × 0.75Z2, then it predictsabsorptions 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
n2E = E/75076 = 0.3 eV
But Kα transition means an energy of
Eα = 10.2(Z − 1)2 eV = 66.9 keV
then∆E
Eα= 4.5× 10−4%
1.8 Radiative life time
Eq (1.10) provides the power of dipole radiation which is rate of changeof energy, for the circular motion, the power doubles since a circular motioncan be decomposed into two linear oscillations,
τ = E/P = ~ω12πǫ0c
3
2e2r2ω4∝ 1/(er)2ω3
For photons of wavelength 650 nm, ω = 2.89× 1015 rad/s; let r = a0, thelife time will be 2.7 × 10−7 s.
3
Table 1: Frequency shiftsB [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 = 600nm, ω = 2πc/λ = 3.13×1015 rad/s. There-fore, 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 magneticfield 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 sinusoidalmotions. 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 ob-served and the intensities are Iσ+ = Iσ− = I;
• Perpendicular to the magnetic field, motion along z direction and pro-jected horizontal motions can all be observed, and
• 2Iσ+ = 2Iσ− = Iπ = I since the projected horizontal motions are onlyhalf 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 is2I/2I = 1.
4
1.12 Bohr theory and the correspondence principle
Energy of hydrogen atoms takes the form
E = K + V = −K = −1
2mev
2
But
mev2
r=e2/4πǫ0r2
Then
E = −1
2
e2/4πǫ0r
, dE =e2/4πǫ02r2
dr
Hence
ω = ∆E/~ =e2/4πǫ02r2~
∆r
The angular frequency is also given by
ω2 =v2
r2=e2/4πǫ0mer3
Equating the two expressions of ω, we have
∆r = 2
√
r~2
mee2/4πǫ0= 2
√a0r
which is equivalent to∆r
∆n= 2(a0r)
1/2 .
Approximate the equation above by the corresponding differential equa-tion, namely r′ = 2
√a0r, and the solution turns out to be
r = a0n2. (1.14)
1.13 Rydberg atoms
Eq (1.4), then
dE
dn= −e
2/4πǫ02a0
d (n−2)
dn=e2/4πǫ0a0n3
= Ry/n3
and for n = 50, ∆E = 1.1 × 10−4 eV. The radius of such atoms is around2500a0 or 132 nm according to Eq (1.14).
5
2 The hydrogen atom
Key formulas
Ylm(θ, φ) = (−1)m
√
(2l + 1) (l −m)!
4π(l +m)!eimφPm
l (cos θ) (2.1)
∫ π
−πdθ eimθe−inθ = 2πδmn (2.2)
∫ π
−πdθ
cosmθ cosnθsinmθ sinnθ
= πδmn (2.3)
∫ π
−πdθ sinmθ cosnθ = 0 (2.4)
∫
dΩ =
∫ 2π
0dφ
∫ π
0sin θdθ =
∫ 2π
0dφ
∫ 1
−1d(cos θ) (2.5)
Es-o = β 〈S · L〉 = β
2(j(j + 1)− l(l + 1)− s(s+ 1)) (2.6)
β =~2
2m2ec
2
e2
4πǫ0
1
(na0)3l(1 +12)(l + 1)
(2.7)
L2 = −
1
sin θ
∂
∂θ
(
sin θ∂
∂θ
)
+1
sin2 θ
∂2
∂2φ
(2.8)
2.1 Angular-momentum eigenfunctions
From the table of spherical harmonics and Eq (2.2), (2.5), we have
• 〈 l1m| l2n〉 ≡ 0 if m 6= n, therefore
〈11| 00〉 = 0
And also
〈10| 00〉 = (constant)×∫ 1
−1cos θ d(cos θ) = 0
• For l = 1, 2, we only need to show that 〈10| 20〉 = 〈11| 21〉 = 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),
Yl,l−1 = (−1)l−1
√
2l + 1
4π(2l − 1)ei(l−1)φP l−1
l (cos θ)
6
• It is convenient to write,
〈 l, l − 1| l − 1, l − 1〉 = 〈l, l| (L−)† |l − 1, l − 1〉
= 〈l, l| L+ |l − 1, l − 1〉= 0
2.3 Radial wavefunctions
With n = 2, l = 1 the integral reads,
⟨1
r3
⟩
=
∫ ∞
0
1
r3R2
2,1(r)r2 dr
=
∫ ∞
0
dr
r
(Z
2a0
)5 ( 2√3
)2
r2e−r/a0
= 1/(24a30)
Invoking the rhs formula, namely
⟨1
r3
⟩
=1
l(l + 12)(l + 1)
(Z
na0
)2
(2.9)
yields the same result 1/(24a30).
2.4 Hydrogen
The probability is given by
∫ rb
0r2 dr |ψ(r)|2 =
∫ rb
0
(r/a0)2 d(r/a0)
πe−2r/a0
=
∫ ǫ:=rb
a0
0
y2 dy
πe−2y
≈ ǫ3e−2ǫ/π ∼ (rb/a0)3
The electronic charge density of this region is
ρe ≈ e|ψ(rb/2)|2 =e(1 − rb/a0)
a30π
2.5 Hydrogen: isotope shift, fine structure and Lamb shift
The mass ratio of electron to proton is ∼ 5 × 10−4 and isotope shift isof the same order, namely, if λ = 600nm, order of isotope shift will be
δνisotope ∼ (5× 10−4) · (5× 105 GHz) = 250GHz
7
Relativistic effect is of the order of α2 ≈ 5×10−5, then the wave numberdifference reads
δνfs ∼ (5× 10−5) · (5× 105 GHz) = 25GHz
Lamb shift is 1/10 of the fine structure shift, that is
δνLamb ∼ 2.5GHz
A Fabry-Perot etalon of finesse F = 100, width d = 1cm is supposed toresolve
δν =1
n(2d)F = 0.5× 3× 108 ∼ 0.1GHz (2.10)
but Doppler effect attenuates the resolution to ∼ 0.7GHz ∼ 1GHz.Therefore, isotope shift, fine structure can be resolved but Lamb shift
approaches the limit of the apparatus and hence cannot be accurately ob-served.
2.6 Transitions
From Eq (1.12), we have
Ultraviolet 100 nm 0.45 ns
Infrared 1000 nm 450ns
2.7 Selection rules
Following similar arguments as in Problem 2.1 and notice there is anadditional cos θ in the integrand.
2.8 Spin-orbit interaction
Calculations based on Eq (2.6) give
Ej =β
2l Ej′ = −β
2(l + 1)
and the mean of the two is
E = (2j + 1)Ej + (2j′ + 1)Ej′ = β[(l + 1)l − l(l + 1)] = 0
2.9 Selection rule for the magnetic quantum number
The integral is readily obtained in text of section 2.2.1, and the resultfollows the same arguments of Problem 2.1.
8
2.10 Transitions
(a) The wavefunction takes the form Ψ(t, r, θ) = Aψ1eiE1t/~ + Bψ2e
iE2t/~
with A≫ B,ψ1 = R1,0Y00, ψ2 = R2,1Y1,0
and|Ψ|2 ≈ A2|ψ1|2 + 2|AB||ψ∗
1ψ2| cos(ω12t)
The second term can be written as
f(r)r cos θ cos(ω12t) = f(r)r · z cos(ω12t)
= f(r) z cos(ω12t)
The sketch of the orbital is as follows (Figure 1)
time
Figure 1: Contour of electron density, or the orbital of wavefunctionΨ(t, r, θ) = Aψ1e
iE1t/~ + Bψ2eiE2t/~ with A ≫ B, ψ1 = R1,0Y00, ψ2 =
R2,1Y1,0 during one period of oscillation.
(b) If Π |Ψ〉 = ±1 |Ψ〉, namely the state exhibits parity, then
〈r〉 = 〈Ψ| Π†rΠ |Ψ〉 = −〈r〉
hence 〈r〉 = 0. But the Hamiltonian of hydrogen atom commutes withΠ, then the eigenstates are of specific parities. The only possibility for anone-vanishing 〈r〉 is to require ψ1 and ψ2 to possess opposite parities.
(c) Now set a0 = 1, the radial integral yields
∫ ∞
0
dr√6r4e−
3
2r =
1√6
(2
3
)5
The angular integral is exactly 1. The total electric dipole moment is
−eD = − 1√6
(2
3
)5
ea0 cosωt z
where a0 is put back through dimension analysis.
9
(d) The density distribution should be more or less similar to Figure 1 fora 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 characterizedby theml but the role of ground state 1s is crucial as pointed out in (b). The1s state is an exponential decay which binds the electron to the region aroundnucleus and hence it is the quantum analogue of the classical restoring force−ω2r.
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 toΘ′
Θ= m
cos 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+1Iang = −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)lYlm (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 theyare transitions from 5 to 4, 4 to 3 and 3 to 2 respectively.
10
3 Helium
3.1 Estimate the binding energy of helium
(a) The total Hamiltonian is the sum of two individual and one interactionHamiltonians,
H = H1 + H2 + Hint (3.1)
Hi = − ~2
2m∇2
i −Ze2
ri(3.2)
Hint =e2
|r1 − r2|(3.3)
where e2 = e2/4πǫ0.
(b) The energy
E(r) =~2
2mr2− Ze2
r
assumes minimum at
rm =~2
Zme2
and it is
E(rm) = −Z2me4
2~2
(c) The repulsive energy, namely electron-electron interaction, is
Eint =e2
r12≈ e2
rm=Zme4
~2
The ionization energy of one electron, according to the estimated energies,is Eion = 0 But experiment gives Eion = 24 eV, then the average distancebetween the two electrons should be greater than rm.
(d) For Si12+, Eion = (142 − 14 × 2) × 13.6 = 2285 eV, excluding therepulsion we have E = 2666 eV. Comparing with results of helium, repulsiongets irrelevant for larger Z.
3.2 Direct and exchange integrals for an arbitrary system
(a) The direct and exchange integrals are respectively,
J =
∫
d3x1d3x2 |uα(r1)|2 |uβ(r2)|2
e2
r12(3.4)
K =
∫
d3x1d3x2 u
∗α(r1)u
∗β(r2)
e2
r12uα(r1)uβ(r2) (3.5)
11
Yet, |A〉 = (2)−1/2(|αβ〉 − |βα〉),
〈A| H ′ |A〉 = 1
2(〈αβ| − 〈βα|)H ′(|αβ〉 − |βα〉) (3.6)
=〈αβH ′αβ〉 − 〈αβH ′βα〉+ α↔ β
2(3.7)
Note that⟨αβH ′αβ
⟩=
⟨βαH ′βα
⟩
⟨αβH ′βα
⟩=
⟨βαH ′αβ
⟩∗
Therefore, it holds for real-valued u(r) that
〈A| H ′ |A〉 =⟨αβH ′αβ
⟩−
⟨αβH ′βα
⟩= J −K
(b) The symmetric wavefunction is construct as
|S〉 = (2)−1/2(|αβ〉+ |βα〉)
The inner product reads
〈A|S〉 = (1− 1 + 〈αβ| βα〉 − c.c)
2= Im 〈αβ| βα〉
Again, for real u(r), 〈A|S〉 are orthogonal.
(c) Since H ′ is invariant under the interchange of particle labels, let Σdenote such an operation, then
〈A| H ′ |S〉 = 〈A|Σ†H ′Σ |S〉 = −〈A| H ′ |S〉
Hence 〈A| H ′ |S〉 = 0.
3.3 Exchange integrals for a delta-function interaction
(a) The Hamiltonian is simply the kinetic energy,
H = − ~2
2m∂2x
For the fact that
u′′0 = −(π/l)2u0 u′′1 = −(2π/l)2u1,
then the energies are
E0 =~2π2
2ml2E1 =
2~2π2
ml2
12
(b) The direct integral is
J =
(2
l
)2 ∫
dx1dx2 sin2(πx1l
)aδx1−x2sin2(
πx2l
)
= a
(2
l
)2 ∫ l
0sin4(πx/l)dx
= a4
πl
∫ π
0sin4 x′dx′ =
3a
2l
The exchange integral is the same thing, K = J . Then the energy shift willbe 3a/l. There is only one state, the symmetric one.
(c) All the eigenstates takes the form,
un(x) =
√
2
lsin
nπx
l
and trivially Jmn = Kmn since the delta potential identifies the two coordi-nates. 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’, otherwisethere is no interaction, and hence no energy shift!
(d) See Figure 2.
0.40.60.8
-0.8-0.6-0.4-0.2
00.2
-1
1
-0.6
0.8
-0.2
0.610.8
0.2
x10.40.6
0.4
0.6
0.20.2x2
00
1
Figure 2: The horizontal axes are the coordinates x1 and x2, and the verticalaxis marks the values of function ψ = u1(x1)u2(x2).
(e) Possible total spins are 0 and 1. The symmetric spatial functionscorresponds to 0-spin, and the antisymmetric spatial functions should havespin-1 or spin-0 but they do not exist.
13
(f) The wavefunctions are independent of particle masses, therefore all themathematics are invariant once the states are given. The energy levels arestill
E+ = E1 +E2 + 2J E− = E1 + E2
but the difference is that the levels are not related to the exchange symme-tries of the particles.
3.4 A helium-like system with non-identical particles
The exoticon-exoticon system (identical fermions) is indifferent with theelectron-electron system except for richer spin configurations for the twosymmetries. Yet no restriction upon symmetries of particle exchange isplaced on exoticon-electron system, therefore, the spatial orbitals can befreely occupied by the two fermions, and for the fact that spin is not includedin the Hamiltonian, hence the energies levels are left unaltered.
3.5 Integrals in helium
Set a0 = 2Z, the integral in the curly brackets is
−r2e−2r2/2− e−2r2/4 + 1/4
and
J =e2
2a0
5
4Z = 34 eV
3.6 Calculation of integrals for 1s2p configuration
-1
0
1
2
3
4
5
6
0 2 4 6 8 10
Rnl
r / a0
R10R20R21
Figure 3: The plot of R10, R20 and R21 with Z = 2.
The integral
J1s2p = − e2
2a0× 0.00208 = 0.0283 eV
14
3.7 Expansion of 1/r12
Expansion in terms of spherical harmonics is the following
1
r12=
1
r2
∞∑
k=0
(r1r2
)k 4π
2k + 1
k∑
q=−k
Y ∗kq(θ1, φ1)Ykq(θ2, φ2) (3.8)
(a) Setting k = 0, 1, we have
1
r12≈ 1
r24π/4π
+r1r2
4π
3
3
4π(cos θ1 cos θ2 + sin θ1 sin θ2 cos(φ1 + φ2))
=1
r2
(
1 +r1r2
cos θ12
)
(b) Mathematically, the expectation values of 1/r12 are always sandwichedby ‘bra-ket’ where the phase factors cancels; this is seen in the expressionfor K1s nl. Physically, the quantum number m is responsible for magneticinteractions but Coulomb repulsions have no interests in that.
(c) Due to the orthogonality relations of spherical harmonics, the termsin 1/r12 are eliminated excluding the one with quantum numbers lm. Forl = 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 theorder k in Eq (3.8).
4 The alkalis
4.1 Configuration of the electrons in francium
Fr = [Rn]7s1
Rn = [Xe]5d105f146s26p66d10
Xe = 1s22s22p63s23p63d104s24p64d104f145s25p6
4.2 Finding the series limit for sodium
Energy levels of sodium obey the rule
E = − Ry
(n∗)2= − 13.6 eV
(n− δ)2(4.1)
15
Then∆E ∝ E3/2 (∆E)2 ∝ E3
The topmost level must have ∆E = 0, and that energy corresponds ioniza-tion energy which was found to be 5.1 eV. The effective principal numbern∗ =
√
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 defectfor 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
∆E = 13.6
(1
(5− 3.19)2− 1
(7− 3.19)2
)
= 3.22 eV
Then the wavelength of the two identical photons are
λ =hc
∆E/2= 771nm
4.5 Application of quantum defects to helium and helium-
like ions
Throught a direct calculation, the wavelength from 1s3d to 1s2p is ob-tained as 625 nm, compared with 656 nm of hydrogen Balmer-α.
Quantum defects can be evaluated by Eq (4.1), they are listed in Table2. It is readily seen that δs > δp > δd.
Table 2: Quantum defects of helium1s2s 1s2p 1s3s 1s3p 1s3d
0.2356 0.0275 0.2265 0.0287 0.0026
To estimate binding energy of 1s4l states, assume the quantum defectsfor 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 +Z2Ry
42= 75.64 eV
where Z = 2.
4.6 Quantum defects and fine structure of potassium
The wave numbers can be put into three groups (3 row vectors), andtake average and difference within each group. The data set becomes ν =(〈ν〉 , δν). The energies are calculated through E = hc/λ, and the dataset for energies reads E = (〈E〉 , δE). Given that the ionization energy ofpotassium IE = 4.34 eV, the energy levels are E− IE from with we can findout the effective principal quantum numbers and quantum defects. Theyare n∗ = 2.23, 3.26, 4.28 and δl = 1.77, 1.73, 1.72; the pattern confirms thatthey corresponds to 4p, 5p, 6p. The fine structure splitting δE/E dividedby α2 are, for 4p, 5p, 6p, 49.18, 33.45, 26.38.
According to Lande formula,
∆Efs =Z2i Z
2o
(n∗)3l(l + 1)α2Ry (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 wavelengths321.84 nm, 321.93 nm.
4.7 The Z-scaling of fine structure
From Eq (4.2), fine structure splitting of ions scales as Z4, then forNa+10, it is 1.3 × 10−5 × 114 = 0.19 eV; for neutral ions it scales as Z2,namely, for sodium atom it should be 0.0016 eV.
4.8 Relative intensities of fine-structure components
(a) In this problem J = 1/2, 3/2, 5/2, hence the weight ratio should be2 : 4 : 6 = 5 : 10 : 15 = 5 : (1 + 9) : (1 + 9 + 5).
(b) Possible electrical dipole transitions areD5/2−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∑l
m=−l |Ylm(θ, φ)|2 is spherically symmetric. For l = 1, thesum reads
s1 =3
4π(cos2 θ + sin2 θ) =
3
4π
17
Associated Legendre polynomials are defined as follows
Pml (x) = (−1)m(1− x2)m/2
(d
dx
)m
Pl(x) (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 symmetryon electron distributions where electrons are seated in several sphericallysymmetric layers. The central field Hamiltonian is the sum of individualelectron Hamiltonian and hence the Shrodinger equations are decoupled.The arrangements of pseudo-independent electrons constitute the electronconfigurations. (a) The residual electrostatic interactions couples the elec-trons and give rise to splitting of energy levels. (b) Magnetic spin-orbitinteractions include another degree of freedom, the spin, which ‘rotates’ theoriginal eigenstates resulting in further splittings.
5.2 Fine structure in the LS-coupling scheme
The spin-orbit Hamiltonian reads
Hs−o = β1〈s1 · S〉S(S + 1)
S · 〈l1 · L〉L(L+ 1)
+ β2〈s2 · S〉S(S + 1)
S · 〈l2 · L〉L(L+ 1)
:= βLSS · L
(5.1)For 3s4p 3P configuration, S = 1 = L, therefore the relation βLS = β4p/2holds if
β4p =1
2
2∑
i=1
βi 〈si · S〉 〈li · L〉
but consider the symmetry of electrons, the right hand side is just
∑
i
βisi‖li‖ = β4p
5.3 The LS-coupling scheme and the interval rule in calcium
The ground configuration of calcium: 1s22s22p63s23p64s2.The triple lines comes from 3P term with J = 0, 1, 2 with and interval
ratio ∼ 2. The first three of muiltiplet of six lines come term 3DJ=1,2,3 withan interval ratio ∼ 1.5; the rest must be 3P, 3D and 3F from selection rules.
18
5.4 The LS-coupling scheme in zinc
The zero comes from 1S0, then it leaps to triplet 4s4p3P0,1,2, next itjumps to singlet 4s4p1P1 and 4s5s 3S1,
1S0.
5.5 The LS-coupling scheme
The interval rule accurately indicates that the four levels of Mg are 3P0,1,2
and 1P1 of 3s3p. Heavier Fe14+ shows a weaker interval pattern (2.4 ratherthan 2) due to intercombination and therefore ∆S 6= 0 transitions occurs inFe14+ 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 3D and 3S can be directly eliminated pre-serving 1D (5-fold) and 1S (1-fold). Then what’s left is 15 − 5 − 1 = 9 and3P (9-fold) fits.
(b) The first 3 lines are the spin-orbit splittings of 3P term, the lattertwo are 1D and 1S. The weak emissions lines near 1D indicates deviationsfrom LS-coupling scheme.
(c) In order to make MS = 2 and ML = 2 for six d-electrons, ms =(12 ,
12 ,
12 ,
12 ,
12 ,−1
2) and ml = (2, 05) where curly bracket means ‘all combi-nations.’ To meet Pauli exclusion principle and maximize MS , the longestsequence of aligned ms is up to 5 since ld = 2, namely, five electrons areequally spread out, therefore ms takes the form above and ML = ml(ms =12) ≤ ld = 2.
5.7 Transition from LS- to jj-coupling
For configuration 3p4s, Hre > Hs-o due to a higher correlation energy (ex-change integral) from inner core, while 3p7s is just the opposite, the electronsseems more independent. Therefore, LS-coupling is proper for 3p4s and jj-coupling for 3p7s. The former gives two terms 3P and 1P while the lattergives 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 5.10in 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 taketheir ratios, I find ∆02/∆11 = 0.23, 0.70, 0.89. As n grows, the LS-couplingtransitions to jj-coupling, namely ∆02/∆11 → 1.
The g-factor gJ = 1.06 should belong to 1P (spin-0), the deviation from1 is due to the mixing with other 3P 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 = 2.4d95s5p is the candidate that mixes with 4d105p.
5.10 The anomalous Zeeman effect
Electric dipole transition rules require ∆MJ = 0,±1. The energy isgiven by
EZM = gJµBBMJ (5.2)
The transitions 3S1(gJ = 2)-3P2(gJ = 32) take place at MJ : [1, 0,−1] →
[2, 1, 0,−1,−2], then (gJMJ) : [2, 0,−2] → [3, 32 , 0,−32 ,−3]. Therefore,
∆(gJMJ) = [(1, 32 , 2); (−12 , 0,
12); (−2,−3
2 ,−1)] The spacing is µBB/2 =14× 1/2 = 7GHz with B = 1T
5.11 The anomalous Zeeman effect
Formula for the g-factor:
gJ =3
2+S(S + 1)− L(L+ 1)
2J(J + 1)(5.3)
then gJ [3S1] = 2 and gJ [
3P1] =32 . 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 Zeemansplitting, possible transitions are:
3P1 → 3D2 (1),3P2 → 3D1 (2)
and the g-factors are
3P1 = 3/2 3P2 = 3/23D1 = 1/2 3D2 = 7/6
The quantities
(gJMJ)1 : [3/2, 0,−3/2] → [7/3, 7/6, 0,−7/6,−7/3]
and(gJMJ)2 : [3, 3/2, 0,−3/2,−3] → [1/2, 0,−1/2]
Therefore, 6∆(gJMJ)1 = [+9,+7,+5;+2, 0,−2;−5,−7,−9] fits the data.
20
5.13 The anomalous Zeeman effect in alkalis
(a) gJ [(2S1/2,
2P1/2,2P3/2)] = (2, 2/3, 4/3)
(b) The g-factors are
gJ [3s2S1/2] = 2 gJ [3p
2P3/2] = 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 2/3µBB = 2/3 × 14× 1 = 9.33GHz(c) There are only two lines of same intensity, no ∆MJ = 0 transition.(d) The energy of fine structure is 517.96GHz this corresponds to a huge
magnetic field 517.96/9.33 = 55.5T.
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
Be = −2
3µ0gsµB|ψns(0)|2s (6.1)
where
|ψns(0)|2 =Z3
πa30n3
Setting Z = 1, s = 1/2 and n = 1, 2 for 1s and 2s, we have magnetic fieldflux density at center as 16.7T and 2.1T respectively. The magnetic fieldfelt by an orbiting electron is B = βl where β is the spin-orbit interactionconstant given by Eq (2.7). Then the field that 2p-electron in hydrogenexperiences is 0.2T.
6.2 Hyperfine structure of lithium
Hyperfine and fine structure splittings are both proportional to the mag-netic moment (spin) and in turn, proportional to the mass. Therefore, hy-perfine structure is of order me/Mp smaller than fine structure.
Let P = maxI, J and Q = minI, 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 from2p 2P3/2,1/2 with maximum J = 3/2 on the top. For 6Li (at J = 3/2),2I + 1 = 3, then I6 = 1 and for 7Li, 2I + 1 = 4, I7 = 3/2, and notice bothare 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 ofnuclear 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 thegap ratio will be 3 : 4. From the 2s hyperfine structures, we can find out thegI 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 = 91.9MHz
6.3 Hyperfine structure of light elements
The hyperfine structure can be estimated as
EHFS ∼ ZiZ2o
(n∗)3me
Mpα2Ry (6.2)
For ground state hydrogen, the hyperfine structure should be around 95MHzand for ground state lithium (Zi = 3, Zo = 1, n∗ = 1.59), 71MHz. Bothstructures are badly underestimated.
6.4 Ratio of hyperfine splittings
For 1s state, J = 1/2, put in notation element : I;F we have relevantquantum numbers for H, D and 3He+:
H : 1/2; 1, 0 D : 1; 3/2, 1/2 3He+ : 1/2; 1, 0
Given A ∝ gIµNZ3, gI ∝ µI/I and interval rule, we have
EH/ED =AH
3/2AD=
3/2 × 2.79
3/2× 0.857= 4.3
EH/E 3He+ =AH
A 3He+=
2.79
−2.13 × 8= −0.16
6.5 Interval for hyperfine structure
(a) The interaction can be expressed in terms of F, I, J ,
EF = A 〈I · J〉 = A
2F (F + 1)− I(I + 1)− J(J + 1) (6.3)
22
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 bothisotopes, then A( 8S7/2; 153) = 20 × (4.86 − 2.35)/(6.42 − 0.77) = 8.9MHz.
6.6 Interval for hyperfine structure
From the argument in (6.2), I postulate that 2I+1 = 6 and I = 5/2. Thesplittings make the ratio 3.1 : 2.6 : 2.0 : 1.5 : 1.0 consistent with prediction3.0 : 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 that70 and 42 are two levels of 39K and 5 and 3 the corresponding two levelsof 40K. Since the intensities depend on degeneracies 2F + 1, then the twolevels 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 = gII, 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 fromthe 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 = gFµBBMF (6.4)
Same separations is equivalent to same gF which is
gF =〈J · F 〉F (F + 1)
gJ =F (F + 1) + J(J + 1)− I(I + 1)
2F (F + 1)gJ (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 = gJµBBMJ +AMIMJ (6.6)
If µBB ≫ A, and J is same for both hyperfine states, then Zeeman splittingsare the same.
(e) The cross-over can be defined at µBB = A = 3.4GHz, or B =3.4/14 = 0.24T.
23
6.9 Isotope shift
The mass effect shifts the energy by
∆νMass ≈me
Mp
δA
A′A′′ν∞ (6.7)
while volume effect gives
∆νVol ≈⟨r2N
⟩
a20
δA
A
Z2
(n∗)3R∞ (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 = .777m−1 (b) .202 + .590 =.792m−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 ≈1MHz. 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 =me
Mpν∞R∞/a
30 × 32/(1.2 × 10−15)2 A = 71
6.12 Specific mass shift
Substitute the nucleus momentum for pN = −∑N
i pi, then kinetic en-ergy becomes
T =1
2me
(
1 +me
MN
) N∑
i=1
p2i
︸ ︷︷ ︸
normal mass effect
+ 2
N∑
i<j
pi · pj
︸ ︷︷ ︸
specific mass effect
24
6.13 Muonic atom
The radius is easily found to be
r =n2~2
Zmµe2= a0/2277
since n = 1, mµ = 207me and Z = 11. The ‘muonic Bohr radius’ is waymuch smaller than electronic Bohr radius. The binding energy is
En = −Z2e4mµ
2n2~2= −112 × 207Ry/n2
If n = 1, E1 = −25047Ry. Volume effect must be extremely evident sincethe size of orbit is comparable to nucleus scale. Actually, it is about 4% ofthe transition energy.
7 The interaction of atoms with radiation
7.1 Averaging over spatial orientations of the atom
(a) We need to show that | 〈x〉 |2 = | 〈r〉 |2/3.
Proof.
∫
dΩ (cosφ sin θ)2 =
∫ 1
−1d(cos θ) (1− cos2 θ)
∫ 2π
0dφ
1 + cos(2φ)
2
=
[
y − y3
3
]1
−1
× 2π
2
=4π
3=
1
3
∫
dΩ.
(b) Since there is no specific preference of coordinate system, the resultin (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 constantphases, without loss of generality, set
c2 =|Ω|W
sin
(Wt
2
)
= const× (eiWt/2 − e−iWt/2)
and plug it into the equation 7.26
c2 + i(ω − ω0)c2 +
∣∣∣∣
Ω
2
∣∣∣∣
2
c2 = 0 (7.1)
25
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.
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
-2 0 2 4 6 8 10 12 14
-2 0 2 4 6 8 10 12 14
F1
F2
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 initialconditions c(0) = [1, 0],
c1(t) = cos(Ωt/2) c2(2) = sin(Ωt/2)
(b) A general solution takes the form,
|ψ〉 = cos
(Ωt
2+ φ
)
|1〉 − i sin
(Ωt
2+ φ
)
|2〉 (7.3)
Initially,|ψ(0)〉 = cosφ |1〉 − i sinφ |2〉
26
Now set Ωt = π, then
|ψ(tπ)〉 = − sinφ |1〉 − i cosφ |2〉 = −icosφ |2〉 − i sinφ |1〉
(c) Set Ωt = 2π, |ψ(t2π)〉 = − |ψ(0)〉.(d) Set φ = 0 and Ωt = π/2,
∣∣φ(tπ/2)
⟩= (|1〉 − i |2〉)/
√2
(e) First set φ = 0, Ωt = π/2, then the state becomes
∣∣1′
⟩= |1〉 − i |2〉 eiϕ/
√2
with an orthogonal state
∣∣2′
⟩= −ie−iϕ |1〉+ |2〉/
√2.
Here comes a second π/2-pulse, the state evolves into
1− e−iϕ
2|1〉 − i
1 + eiϕ
2|2〉
Therefore, the probabilities in |1〉 and |2〉 are sin2(ϕ/2) and cos2(ϕ/2).(f) Since cos(φ+ π/4) = (cosφ− sinφ)/
√2 and sin(φ+ π/4) = (cos φ+
sinφ)/√2, then
Uπ
2=
1√2
(1 −i−i 1
)
Similarly,
Uπ =
(0 −i−i 0
)
and the phase shift in |2〉Uϕ =
(1 00 eiϕ
)
Then the successive operator reads
U = Uπ
2UϕUπUπ
2
= −eiϕ/2(
cos ϕ2 sin ϕ
2− sin ϕ
2 cos ϕ2
)
7.4 The steady-state excitation rate with radiative broaden-
ing
(a) The differential equation reads c2 = −Γ2 c2, the solution reads, from
inspection, c2(t) = c2(0)e−Γt/2.
27
(b) Sinced
dt
(
c2eΓt/2
)
= −ic1Ω∗
2e−i(ω−ω0+iΓ/2)t
integration gives
c2(eΓt/2 − 1) = −iΩ
∗
2
e−i(ω−ω0+iΓ/2)t
Γ/2− i(ω − ω0)
where c1 = 1 in the weak excitation. In the limit Γt≫ 1,
|c2|2 =Ω2/4
(ω − ω0)2 + Γ2/4
7.5 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−Nσ(ω)z (7.4)
For transmission 1/e, Nσ = 1. The absorption cross-section reads
σ(ω) =g2g1
π2c2
ω20
A21gH(ω) (7.5)
where g1, g2 are the degeneracies of two states and gH is the Lorentzian lineshape function
gH(ω) =1
2π
Γ
(ω − ω0)2 + Γ2/4(7.6)
For a simple two level system (Γ = A21) with g2/g1 = 3 at resonance,
σ(ω0) = 3× 2πc2
ω20
A21
Γ=
3λ202π
(7.7)
Sodium s-p transition has λ0 = 589nm, invoking Eq (7.7), we haveN = 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 = ω0σ(ω0)/ωσ(ω). The absorptioncoefficient is defined
κ(ω, I) =Nσ(ω)
1 + I/Is(ω)(7.9)
then we can find κ(ω, Isat). Notice τ = 1/Γ = 6ns for sodium at λ = 589nm.
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 and6.5× 107 s−1, the lifetimes τ = 1/A21 are then 160, 5.5, 15 ns.
(c) Shorter lifetime means larger spontaneous transition rate. Since 1s-2p overlaps much more than 1s-3p, transition rate is higher for the formerand hence shorter lifetime.
(d) The Einstein coefficients satisfy the following relations
A21 =~ω3
π2c3B21 g1B12 = g2B21 (7.10)
and radial matrix element D12 is contained in
B12 =πe2|D12|23ǫ0~2
⇒ A21 =g1g2
4α
3c2ω3|D12|2 (7.11)
Then the D12’s for transitions 2p-3s, 1s-3p, 2s-3p, 2p-3d and 1s-2p are0.53, 0.52, 3.0, 3.8, 1.3, measured in a0.
(e) The saturation intensity is given by
Isat =π
3
hc
λ3τ=π
3
hcΓ
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), thereforeIsat = (0.46, 3.7 × 103)mW/cm2.
7.7 The classical model of atomic absorption
The dynamical equation takes the form
x+ βx+ ω20x =
F (t)
mcosωt (7.13)
The steady state ansatz takes the general form x(t) = Aei(ωt+δ) and plugit into the dynamical equation (I have secretly replaced cosωt by eiωt), wehave
−ω2 + iβω + ω20 =
F (t)
mAe−iδ
(a) Take the real part of the ansatz,
x = A cos(ωt+ δ)
= A(cosωt cos δ − sinωt sin δ)
: = u cosωt− v sinωt
29
(b) Notice that
√
u2 + v2 = A
=
∣∣∣∣
F0/m e−iδ
ω2 − ω20 + iβω
∣∣∣∣
=F0/m
√
(ω2 − ω20)
2 + (βω)2
≈ F0
2mω
(
(ω − ω0)2 +
β2
4
)−1/2
The peak is located at ω2 = ω20 − β2/2.
(c) The phase is given by
tan δ =v
u
=βω
ω2 − ω20
The first line comes from (a) and the second line comes from the complexalgebraic equation above.
(d) Combine (b) and (c).(e) Notice that P ∝ |A|2 and simply invoke (b) we have,
P ∝ 1
(ω − ω0)2 + (β/2)2
7.8 Oscillator strength
(a) Integrate the cross-section in Eq (7.5) gives
∫ ∞
−∞σ(ω)dω =
g2g1
π2c2
ω20
g1g2
4α
3c2|D12|2
⟨ω3
⟩
≈ π2c2
ω20
4α
3c2|D12|2ω3
0
= 2π2r0cf12
where, as you may check, r0 = α~/mec = 2.8 × 10−15. Note, I computedthe complicated integral with computer algebra system and made approxi-mations by hand.
(b) The cross-section can be calculated from
σ(ω) =P (ω)
(c/8π)E20
=e4ω4
3m23c
3
8π/c
(ω2 − ω20)
2 + β2ω2
30
(c) For an harmonic oscillator E = K + U = 2K, if we assume thatoscillation is much faster than the damping, then this formula holds and wehave
E = m⟨x2
⟩= mx20e
−βtω′4⟨cos2(ω′t)
⟩= mx20e
−βtω′4/2
Then, according to the classical damping formula (1.23), the damping rateis
1
τ= − E
E= β =
e2ω2
6πǫ0mec3
(d) Combine the result in (b) and (c).(e) From Eq (7.11), we have
|D12|2 =g2g1
3λ3
4αc(2π)3A21 (7.14)
and for sodium 3s-3p transition, A21 = 2π × 107 s−1, λ = 589nm, therefore|D21|2 = 5.3× 10−20 C2m2. Absorption oscillator strength is, by definition,
f12 = 2meω0D212/(3~) = 0.980 (7.15)
(f) For hydrogen 1s-2p and 1s-3p the oscillator strengths follow from theformula
f12 =3
2
mc2A21
~αω2
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 trans-form and finally take the real part. It is straightforward if you look it up inthe table and the final result will be a Lorentzian with peak at ω.
7.11 The balance between absorption and spontaneous emis-
sion
It comes from the Einstein equation
N2 = (N1B12 −N2B21)ρ(ω12)−N2A21 (7.16)
at equilibrium and if we neglect the contribution from spontaneous emission,the absorption is
κI = (N1B12 −N2B21)ρ(ω12)~ω = N2A21~ω
31
Note that N2 = Nρ22 and substitute eqn (7.69) for ρ22, then we have eqn(7.87) invoking the definition
I
Isat=
2Ω2
Γ2(7.17)
7.12 The d.c. Stark effect
(a) The perturbation is similar to “intermediate field strength” in thediscussion of Zeeman effect. The eigenvalues are
λ1,2 = ±√
V 2 + ǫ2/4 ≈ ±(ǫ
2+V 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 =eE0a
20/2 ∼ 10−18 eV. Then ∆E = V 2/ǫ ∼ 10−37 eV.
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