foot atomic solutions by zhao, c

7/25/2019 Foot Atomic Solutions by Zhao, C.

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Solutions to the Exercies ofC. J. Foots Atomic Physics

Chenchao ZhaoDepartment of Physics, Beijing Normal University, Beijing, China

(Dated: June 28, 2011)

1 Early atomic physics

Key formulas:

1

=R

1

n2 1

n2

(1.1)

RH=R mH

me+ mH(1.2)

a0=

2

(Ze2

/40)me= 0.529

1010 m (1.3)

E= Ze2/402a0

1

n2 Z2 (1.4)

E=2

n2E (1.5)

=e2/40

c = 1/137 (1.6)

f

c =R((Z K)2 (Z L)2) (1.7)

() = 3

2c31

exp /kT

1 (1.8)

N2g2

=N1

g1exp

kT

(1.9)

P = e2x20

4

120c3 (1.10)

L= eB

2me(1.11)

=60mec

3

e22 (1.12)

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1.1 Isotope shift

To find wavelengths of Balmer- transitions, we set n = 2 and n = 3 asin Eq (??) Then = 365R , hence

H D = 365

(1/RH 1/RD)

=36me5R

1

mH 1

mD

18me

5mHR= 0.18nm

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

HHe

=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

=

E

E =

4

2 = 75076 (1.13)

for the fact that = ch/E and

d= chdE/E2= (ch/E) (dE/E)= (dE/E)

This corresponds to a grating of 105 grooves.

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1.4 X-rays

Eq (1.7) reduces to f= 3cRZ2/4 when Z . Therefore, f Z

1.5 X-rays

Since E = hf = 3/4(hcR)Z2 = 13.6 eV0.75Z2, then it predicts

absorptions at around 6.4keV and 6.9keV.

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 = 22.9keV

and Eq (1.5) gives

E=2

n2E= E/75076 = 0.3 eV

ButK

transition means an energy of

E= 10.2(Z 1)2 eV = 66.9keV

thenE

E= 4.5 104%

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,

=E/P = 120c

3

2e2r24 1/(er)23

For photons of wavelength 650 nm, = 2.89 1015 rad/s; letr = a0, thelife time will be 2.7 107 s.

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Table 1: Frequency shiftsB [T] 3 105 1L [MHz] 2.6 8.8 104L/(10

14 Hz) 2.6 108 8.8 104

1.9 Black-body radiation

SettingN2= 0.1N, N1 = 0.9N, g1 = 1, g2 = 3, Eq (1.9) gives

exp(/kT) = 27

provided the wavelength = 600nm, = 2c/= 3.13

1015 rad/s. There-

fore, T = 7.23 103 K and Eq (1.8) gives the density= 4.70 1016 J s/m3

1.10 Zeeman effect

The Larmor frequency is given by Eq (1.11), and the Earth magneticfield is about 3 105 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. LetIdenote 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 I+= I= I;

Perpendicular to the magnetic field, motion alongz direction and pro-jected horizontal motions can all be observed, and

2I+= 2I= I = Isince 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.

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1.12 Bohr theory and the correspondence principle

Energy of hydrogen atoms takes the form

E= K+ V = K= 12

mev2

But

mev2

r =

e2/40r2

Then

E= 12

e2/40r

, dE=e2/40

2r2 dr

Hence

= E/=e2/40

2r2 r

The angular frequency is also given by

2 = v2

r2 =

e2/40mer3

Equating the two expressions of, we have

r= 2

r2

mee2/40= 2

a0r

which is equivalent tor

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

2/402a0

d (n2)

dn =

e2/40a0n3

= Ry/n3

and for n = 50, E = 1.1 104 eV. The radius of such atoms is around2500a0 or 132 nm according to Eq (1.14).

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2 The hydrogen atom

Key formulas

Ylm(, ) = (1)m

(2l+ 1) (l m)!4(l+ m)!

eimPml (cos ) (2.1)

d eimein = 2mn (2.2)

d

cos m cos n

sin m sin n

=mn (2.3)

d sin m cos n= 0 (2.4) d =

20

d

0

sin d =

20

d

11

d(cos ) (2.5)

Es-o= S L = 2

(j(j+ 1) l(l+ 1) s(s + 1)) (2.6)

=

2

2m2ec2

e2

40

1

(na0)3l(1 + 12 )(l+ 1)

(2.7)

L2 =

1

sin

sin

+

1

sin2

2

2

(2.8)

2.1 Angular-momentum eigenfunctions

From the table of spherical harmonics and Eq (2.2), (2.5), we have

l1m| l2n 0 ifm =n, therefore11| 00 = 0

And also

10| 00 = (constant) 11

cos d(cos ) = 0

For l = 1, 2, we only need to show that10| 20 = 11| 21 = 0.Through inspections, the integrands as functions of cos are both odd,hence the integrals vanish.

2.2 Angular-momentum eigenfunctions

According to Eq (2.1),

Yl,l1= (1)l1

2l+ 1

4(2l 1) ei(l1)Pl1l (cos )

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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

r3R22,1(r)r

2 dr

=

0

drr

Z2a0

5 23

2r2er/a0

= 1/(24a30)

Invoking the rhs formula, namely1

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 rb0

r2 dr |(r)|2 = rb

0

(r/a0)2 d(r/a0)

e2r/a0

=

:= rba0

0

y2 dy

e2y

3e2/ (rb/a0)3

The electronic charge density of this region is

e e|(rb/2)|2 = e(1 rb/a0)a30

2.5 Hydrogen: isotope shift, fine structure and Lamb shift

The mass ratio of electron to proton is 5 104 and isotope shift isof the same order, namely, if = 600 nm, order of isotope shift will be

isotope (5 104) (5 105 GHz) = 250 GHz

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Relativistic effect is of the order of2 5105, then the wave numberdifference reads

fs (5 105) (5 105 GHz) = 25 GHz

Lamb shift is 1/10 of the fine structure shift, that is

Lamb 2.5GHz

A Fabry-Perot etalon of finesseF= 100, width d = 1 cm is supposed toresolve

= 1

n(2d)F = 0.5 3 108 0.1 GHz (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 0.45ns

Infrared 1000 nm 450 ns

2.7 Selection rules

Following similar arguments as in Problem 2.1 and notice there is anadditional cos in the integrand.

2.8 Spin-orbit interaction

Calculations based on Eq (2.6) give

Ej =

2

l Ej =

2

(l+ 1)

and the mean of the two is

E= (2j+ 1)Ej+ (2j + 1)Ej =[(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.

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2.10 Transitions

(a) The wavefunction takes the form (t,r,) =A1eiE1t/ +B2e

iE2t/

with A B,1 = R1,0Y00, 2 = R2,1Y1,0

and||2 A2|1|2 + 2|AB||12| cos(12t)

The second term can be written as

f(r)r cos cos(12t) = f(r)r z cos(12t)

=f(r)z cos(12t)

The sketch of the orbital is as follows (Figure 1)

time

Figure 1: Contour of electron density, or the orbital of wavefunction(t,r,) = A1e

iE1t/ +B 2eiE2t/ with A

B, 1 = R1,0Y00, 2 =

R2,1Y1,0 during one period of oscillation.

(b) If | = 1 |, namely the state exhibits parity, then

r = | r | = rhencer = 0. But the Hamiltonian of hydrogen atom commutes with, then the eigenstates are of specific parities. The only possibility for anone-vanishingr is to require 1 and 2 to possess opposite parities.

(c) Now seta0= 1, the radial integral yields 0

dr6

r4e3

2r =

16

2

3

5The angular integral is exactly 1. The total electric dipole moment is

eD= 16

2

3

5ea0cos t z

where a0 is put back through dimension analysis.

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(d) The density distribution should be more or less similar to Figure1for

a constant but it becomes apparent if one writes c = t or= t 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 themlbut 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

2r.

2.11 Angular eigenfunctions: Yll

(a) Raising operator is given as

L+= ei (+ i cot ) (2.11)

Then we write(+ i cot )(e

im) = 0

it is equivalent to

=mcos

sin

(b) The solution is = sinm . ApplyingL2 (see (2.8)) yields

L2(eim) =m(m+ 1)eim

2.12 Parity and selection rules

Ifl1 l2 is even, thenIang= (1)l1l2+1Iang= Iang

which implies the integral vanishes.

To have non-vanishing angular integrals, l1 l2 must be odd. But theparity of spherical harmonics is

Ylm = (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, 1.890eV and theyare transitions from 5 to 4, 4 to 3 and 3 to 2 respectively.

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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

2m2i

Ze2

ri(3.2)

Hint= e2

|r1 r2| (3.3)

where e2 =e2/40.

(b) The energy

E(r) =

2

2mr2 Ze

2

r

assumes minimum at

rm=

2

Zme2

and it is

E(rm) = Z2me4

22

(c) The repulsive energy, namely electron-electron interaction, is

Eint= e2

r12 e

2

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 = (142 142)13.6 = 2285eV, excluding the

repulsion we haveE= 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=

d3x1d

3x2|u(r1)|2 |u(r2)|2 e2

r12(3.4)

K=

d3x1d

3x2u(r1)u

(r2)

e2

r12u(r1)u(r2) (3.5)

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Yet,|A = (2)1/2(| |),

A| H |A = 12

(| |)H(| |) (3.6)

=H H +

2 (3.7)

Note that H

=

H

H

=

H

Therefore, it holds for real-valuedu(r) that

A| H |A = H H =J K(b) The symmetric wavefunction is construct as

|S = (2)1/2(| + |)

The inner product reads

A| S =(1 1 + | c.c)2

= Im |

Again, for real u(r),A| S are orthogonal.

(c) Since H is invariant under the interchange of particle labels, let denote such an operation, then

A| H |S = A| H |S = A| H |S

HenceA| H |S = 0.

3.3 Exchange integrals for a delta-function interaction

(a) The Hamiltonian is simply the kinetic energy,

H= 22m

2x

For the fact that

u0 = (/l)2u0 u1 = (2/l)2u1,

then the energies are

E0=

22

2ml2 E1=

222

ml2

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(b) The direct integral is

J=

2

l

2 dx1dx2sin

2(x1

l )ax1x2sin

2(x2

l )

=a

2

l

2 l0

sin4(x/l)dx

=a4

l

0

sin4 xdx =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

l sin

nx

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 Figure2.

0.4

0.6

0.8

-0.8

-0.6

-0.4

-0.2

0

0.2

-1

1

-0.6

0.8

-0.2

0.61

0.8

0.2

x10.40.6

0.4

0.6

0.2

0.2x2

00

1

Figure 2: The horizontal axes are the coordinatesx1and x2, and the verticalaxis marks the values of function = 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.

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(f ) The wavefunctions are independent of particle masses, therefore all the

mathematics 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 is

placed 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

r2e2r2/2 e2r2/4 + 1/4and

J= e2

2a0

5

4Z= 34eV

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 ofR10, R20 and R21 with Z= 2.

The integral

J1s2p= e2

2a0 0.00208 = 0.0283eV

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3.7 Expansion of1/r12

Expansion in terms of spherical harmonics is the following

1

r12=

1

r2

k=0

r1r2

k 42k+ 1

kq=k

Ykq(1, 1)Ykq(2, 2) (3.8)

(a) Setting k = 0, 1, we have

1

r12 1

r2{4/4

+r1r2

4

3

3

4(cos 1 cos 2+ sin 1 sin 2 cos(1+ 2))

= 1

r2

1 +

r1r2

cos 12

(b) Mathematically, the expectation values of 1/r12are always sandwichedby bra-ket where the phase factors cancels; this is seen in the expressionfor K1snl. 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. 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 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 )2 (4.1)

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Then

E E3/2 (E)2 E3The 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 defect

for 8s is still 1.35, the binding energy will be

ENa 8s = 13.6/(8 1.35)2 = 0.31eVwhile for hydrogenEH 8s = 0.21eV.

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 = 3.22eV

Then the wavelength of the two identical photons are

= 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-.

Quantum defects can be evaluated by Eq (4.1), they are listed in Table

2. It is readily seen that s > p> d.

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, 0.85eV

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The second ionization energy of Li+ is

IE2 = (E E1s4f) = 72.24 + Z2Ry

42 = 75.64eV

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 =( , ). The energies are calculated through E = hc/, and the dataset for energies reads E = (E , E). 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 l = 1.77, 1.73, 1.72; the pattern confirms thatthey corresponds to 4p, 5p, 6p. The fine structure splitting E/E dividedby2 are, for 4p, 5p, 6p, 49.18, 33.45, 26.38.

According to Lande formula,

Efs= Z2iZ

2o

(n)3l(l+ 1)2Ry (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 wavelengths321.84nm, 321.93nm.

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 1.3 105 114 = 0.19eV; 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 be

2 : 4 : 6 = 5 : 10 : 15 = 5 : (1 + 9) : (1 + 9 + 5).(b)Possible electrical dipole transitions are D5/2F5/2, D5/2D7/2, 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 thatl

m=l |Ylm(, )|2 is spherically symmetric. For l = 1, thesum reads

s1= 3

4(cos2 + sin2 ) =

3

4

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Associated Legendre polynomials are defined as follows

Pml (x) = (1)m(1 x2)m/2

d

dx

mPl(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

Hso= 1 s1 SS(S+ 1)

S l1 LL(L+ 1)

+ 2s2 S

S(S+ 1)S l2 L

L(L + 1) :=LSS L

(5.1)For 3s4p 3P configuration, S = 1 = L, therefore the relation LS = 4p/2holds if

4p =1

2

2i=1

i si S li L

but consider the symmetry of electrons, the right hand side is just

i

isi

li

= 4p

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.

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5.4 The LS-coupling scheme in zinc

The zero comes from 1S0, then it leaps to triplet 4s4p3P0,1,2, next it

jumps 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 3 P0,1,2and 1P1 of 3s3p. Heavier Fe

14+ shows a weaker interval pattern (2.4 ratherthan 2) due to intercombination and therefore S= 0 transitions occurs inFe14+ but not Mg.

5.6 LS-coupling for configurations with equivalent electrons

(a)Fornp2 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 whats 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 , 12 ) and ml = (2, {0}5) where curly bracket means all combi-

nations. To meet Pauli exclusion principle and maximizeMS, 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-odue 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 forn = 3, 4, 5, then taketheir ratios, I find 02/11 = 0.23, 0.70, 0.89. Asn 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 from

1 is due to the mixing with other 3P wavefunctions.

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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= gJBBMJ (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,

3

2 , 2);(1

2 , 0,1

2 ); (2, 3

2 , 1)] The spacing is BB/2 =14 1/2 = 7 GHz withB = 1 T


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] = 3

2 . Compared with the previous problem(or Figure 5.13), since J= 0, MJ= 0, there are only six lines left.


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= 3/2

3D1= 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.

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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 2/3 BB = 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 517.96 GHz this corresponds to a hugemagnetic 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

Be = 23

0gsB|ns(0)|2s (6.1)

where

|ns(0)|2 = Z3

a30n3

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 2.1 T respectively. The magnetic fieldfelt by an orbiting electron is B =l where is the spin-orbit interaction

constant given by Eq (2.7). Then the field that 2p-electron in hydrogenexperiences is 0.2 T.

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 = 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 ofF.

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Since 2s has no fine structure, the fine structure totally comes from

2p 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 ofJ = 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 thegIratio 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)3meMp

2Ry (6.2)

For ground state hydrogen, the hyperfine structure should be around 95 MHzand 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 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 gINZ3, gI I/Iand interval rule, we have

EH/ED =

AH

3/2 AD =

3/2

2.79

3/2 0.857 = 4.3

EH/E3He+ = AHA 3He+

= 2.79

2.13 8= 0.16

6.5 Interval for hyperfine structure

(a)The interaction can be expressed in terms ofF, I , J ,

EF =A I J = A2

{F(F+ 1) I(I+ 1) J(J+ 1)} (6.3)

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Therefore, EF EF1= 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 isI=Fmax J= 811/2 = 5/2.(c)From the peaks, it can be inferred that nuclear spin is 5/2 for both

isotopes, thenA( 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 andI= 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 justI = 2 1/2 = 3/2,consistent with the number of splittings (I > J). And the magnetic moment(I=gII, Ithe 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=gFBBMF (6.4)

Same separations is equivalent to same gF which is

gF = J FF(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= gJBBMJ+ AMIMJ (6.6)

IfBB A, andJis same for both hyperfine states, then Zeeman splittingsare the same.

(e) The cross-over can be defined at BB = A = 3.4GHz, or B =3.4/14 = 0.24T.

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6.9 Isotope shift

The mass effect shifts the energy by

Mass meMp

A

AA (6.7)

while volume effect gives

Vol

r2N

a20

A

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 =.777m1 (b) .202 + .590 =.792m1.

6.10 Volume shift

From Eq (6.8), the relative uncertainty that rN contributes is

[Vol] = 2[rN], d[Vol] = 2% Vol

For hydrogen 1s configuration, the volume shift is about 5109 eV1 MHz. Therefore, d[EVol]/ELamb 1% 1/1057.8 1ppm.

6.11 Isotope shift

Assume Z=A/2, n = 2, = 1/500 nm1, then

A11/3 = meMp

R/a30 32/(1.2 1015)2 A= 71

6.12 Specific mass shift

Substitute the nucleus momentum for pN =Ni pi, then kinetic en-ergy becomes

T = 1

2me

1 + meMN

Ni=1

p2i normal mass effect

+ 2Ni


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6.13 Muonic atom

The radius is easily found to be

r= n22

Zme2 =a0/2277

since n = 1, m = 207me and Z = 11. The muonic Bohr radius is waymuch smaller than electronic Bohr radius. The binding energy is

En= Z2e4m

2n22 = 112 207Ry/n2

Ifn = 1, E1 =25047 Ry. 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.

d (cos sin )2 = 1

1 d(cos ) (1 cos2

) 2

0 d

1 + cos(2)

2

=

y y

3

3

11

22

=4

3 =

1

3

d.

(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=||

W sin

W t

2

= const (eiWt/2 eiWt/2)

and plug it into the equation 7.26

c2+ i( 0)c2+

2

2

c2 = 0 (7.1)

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The first exponential gives

(W/2)2 ( 0)W/2 + |/2|2 = 0

the second exponential (with minus sign) gives

(W/2)2 + ( 0)W/2 |/2|2 = 0

Both cases are consistent, if 0, with definition ofW,

W2 = 2 + ( 0)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 and F2 for 3

7.3 - and /2-pulses

(a) Try the ansatz c(t) = c exp(it/2), I find = , then with initial

conditions c

(0) = [1, 0],

c1(t) = cos(t/2) c2(2) = sin(t/2)

(b) A general solution takes the form,

| = cos

t

2 +

|1 i sin

t

2 +

|2 (7.3)

Initially,

|(0) = cos |1 i sin |2

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Now set t= , then

|(t) = sin |1 i cos |2 = i{cos |2 i sin |1}

(c)Set t= 2,|(t2) = |(0).(d) Set = 0 and t= /2,(t/2) = (|1 i |2)/2(e)First set= 0, t= /2, then the state becomes

1

= {|1 i |2 ei}/

2

with an orthogonal state2 = {iei |1 + |2}/2.Here comes a second /2-pulse, the state evolves into

1 ei2

|1 i 1 + ei

2 |2

Therefore, the probabilities in|1 and|2 are sin2(/2) and cos2(/2).(f )Since cos( + /4) = (cos sin )/2 and sin( + /4) = (cos +

sin )/

2, then

U2

= 12

1 ii 1 Similarly,

U =

0 ii 0

and the phase shift in|2

U=

1 00 ei

Then the successive operator reads

U= U2 UUU2

= ei/2

cos 2 sin2

sin 2 cos 2

7.4 The steady-state excitation rate with radiative broaden-

ing

(a) The differential equation reads c2 = 2 c2, the solution reads, frominspection, c2(t) =c2(0)e

t/2.

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(b) Since

ddtc2et/2 = ic1

2 ei(0+i/2)t

integration gives

c2(et/2 1) = i

2

ei(0+i/2)t

/2 i( 0)where c1 = 1 in the weak excitation. In the limit t 1,

|c2|2 = 2/4

( 0)2 + 2/4

7.5 Saturation of absorption

(a)Transmission in a weak resonant field takes the simple form

I(, z) =I(, 0)e()z =I(, 0)eN()z (7.4)

For transmission 1/e, N = 1. The absorption cross-section reads

() = g2g1

2c2

20A21gH() (7.5)

whereg1, g2 are the degeneracies of two states and gHis the Lorentzian line

shape function

gH() = 1

2

( 0)2 + 2/4 (7.6)

For a simple two level system ( =A21) with g2/g1 = 3 at resonance,

(0) = 3 2c2

20

A21

=320

2 (7.7)

Sodium s-p transition has 0 = 589 nm, invoking Eq (7.7), we haveN= 6 1012.

(b) The saturation intensity is defined as

Is() = A212()

(7.8)

and the fact that Isat = Is(0), Isat/Is = 0(0)/(). The absorptioncoefficient is defined

(, I) = N ()

1 + I/Is() (7.9)

then we can find(, Isat). Notice= 1/ = 6 ns for sodium at= 589nm.

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7.6 The properties of some transitions in hydrogen

(b)The A21 for 3s, 3p and 3d are 6.3 106 s1, (17+2.2) 107 s1 and6.5 107 s1, the lifetimes = 1/A21 are then 160, 5.5, 15ns.

(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= 3

2c3B21 g1B12= g2B21 (7.10)

and radial matrix element D12 is contained in

B12= e2|D12|2

302 A21= g1

g2

4

3c23|D12|2 (7.11)

Then the D12s 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=

3

hc

3 =

3

hc

3 (7.12)

For the case where spin and fine structures are ignored, =A21.Now calculate the saturation intensities for 2p-3s and 1s-3p. The A21are

(6.3 106, 1.7 108) s1, 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 + x+ 20x=F(t)

m cos t (7.13)

The steady state ansatz takes the general form x(t) = Aei(t+) and plugit into the dynamical equation (I have secretly replaced cos t byeit), wehave

2 + i + 20 =F(t)

mAei

(a)Take the real part of the ansatz,

x= A cos(t + )

=A(cos t cos sin t sin ): =u cos t v sin t

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(b) Notice thatu2 + v2 =A

=

F0/mei2 20+ i

= F0/m

(2 20)2 + ()2

F02m

( 0)2 +

2

4

1/2The peak is located at 2 =20 2/2.

(c)The phase is given by

tan = v

u

=

2 20The 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( 0)2 + (/2)2

7.8 Oscillator strength

(a)Integrate the cross-section in Eq (7.5) gives

()d =g2g1

2c2

20

g1g2

4

3c2|D12|2

3

2c2

20

4

3c2|D12|230

= 22r0cf12

where, as you may check, r0 = /mec = 2.8 1015. Note, I computedthe complicated integral with computer algebra system and made approxi-mations by hand.

(b) The cross-section can be calculated from

() = P()

(c/8)E20=

e44

3m23c3

8/c

(2 20)2 + 22

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(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 wehave

E= m

x2

=mx20et4

cos2(t)

=mx20e

t4/2

Then, according to the classical damping formula (1.23), the damping rateis

1

= E

E ==

e22

60mec3

(d) Combine the result in (b) and (c).(e)From Eq (7.11), we have

|D12|2

=

g2

g1

33

4c(2)3 A21 (7.14)

and for sodium 3s-3p transition, A21 = 2 107 s1, = 589 nm, therefore|D21|2 = 5.3 1020 C2 m2. Absorption oscillator strength is, by definition,

f12= 2me0D212/(3) = 0.980 (7.15)

(f )For hydrogen 1s-2p and 1s-3p the oscillator strengths follow from theformula

f12=3

2

mc2A212

and they are 0.418, 0.0801.

7.9 Doppler broadening

In the rest frame and moving frame the frequency differs by= 0=kv. And the sinc2 behaves like a delta, then v= (0 )/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 .

7.11 The balance between absorption and spontaneous emis-

sion

It comes from the Einstein equation

N2= (N1B12 N2B21)(12) N2A21 (7.16)at equilibrium and if we neglect the contribution from spontaneous emission,the absorption is

I= (N1B12 N2B21)(12) = N2A21

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Note that N2 = N 22 and substitute eqn (7.69) for 22, then we have eqn

(7.87) invoking the definition

I

Isat=

22

2 (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

1,2= V2 + 2/4 (

2+

V2

)

(b) See part (a).(c) For 3s-3p transition, = hc/ = 2.11 eVThe displacement will be

of the order a0, then the electrostatic energy will be of the order V =eE0a

20/2 1018 eV. Then E= V2/ 1037 eV.

32

foot atomic solutions by zhao, c

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