Atomic, molecular and laser physics

PHY-331 Solution Set # 7 November 29, 2010

Zeeman effect and hyperfine interaction

Solution 1. (a) Weak field

The energies of hydrogen’s n = 3 levels, when placed in a weak magnetic field, are shown
in the Table. The correction to energy in a weak magnetic field is,

∆E = gJ µB mj Bext

where,

j(j + 1) + 3/4 − l(l + 1)

gJ = 1 +
2j(j + 1)
e~
µB = ·
2me

The energy corrections are relative to the respective fine-structure levels (defined by n, l, j)
as shown in Figure 1.

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(5/2,5/2) = 3µΒΒ
(5/2,3/2) = 9/5 µΒextΒext
(5/2,1/2) = 3/5 µΒΒext
3d5/2 (5/2,-1/2) = −3/5 µΒΒext
(5/2,-3/2) = −9/5 µΒΒext
(5/2,-5/2) = 3µΒΒext
(3/2,3/2) = 2µΒΒext (3/2,3/2) = 6/5 µΒΒext
(3/2,1/2) = 2/3µΒΒ (3/2,1/2) = 2/5 µΒΒext
ext
(3/2,-1/2) = −2/3µΒΒ (3/2,-1/2) = −2/5 µΒΒext
3p3/2 (3/2,-3/2) = −2µΒΒ
ext 3d3/2
ext
(3/2,-3/2) = −6/5 µΒΒext
(1/2,1/2) = µΒΒ ext = 1/3µΒΒext
(1/2,1/2)
3s1/2 (1/2,-1/2) =−µΒΒext
3p1/2 (1/2,-1/2) = −1/3µΒΒext

(3/2,3/2) = 2µΒΒext
(3/2,1/2) = 2/3µΒΒext
(3/2,-1/2)
2p3/2 (3/2,-3/2)
= −2/3µΒΒext
= −2µΒΒext

(1/2,1/2) = µΒΒext (1/2,1/2) = 1/3µΒΒext

2s1/2 (1/2,-1/2) = −µΒΒext

2p1/2 (1/2,-1/2) = −1/3µΒΒext

(1/2,1/2) = µΒΒ ext

1s1/2 (1/2,-1/2)
Ε=−µΒΒext

FIG. 1: Splitting of hydrogen’s n = 3 energy levels when placed in a weak magnetic field. Above
each state, the values of j, mj are specified explicitly.

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n l j mj gJ ∆Eweak
3 0 1/2 −1/2 2 −µB Bext
3 0 1/2 +1/2 2 +µB Bext
3 1 1/2 −1/2 2/3 − µB B3 ext
3 1 1/2 +1/2 2/3 + µB B3 ext
3 1 3/2 −3/2 4 −2µB Bext
3 1 3/2 −1/2 4/3 − 2µB3Bext
3 1 3/2 +1/2 4/3 + 2µB3Bext
3 1 3/2 +3/2 4 +2µB Bext
3 2 3/2 −3/2 12/5 − 6µB5Bext
3 2 3/2 −1/2 4/5 − 2µB5Bext
3 2 3/2 +1/2 4/5 + 2µB5Bext
3 2 3/2 +3/2 12/5 + 6µB5Bext
3 2 5/2 −5/2 6 −3µB Bext
3 2 5/2 −3/2 18/5 − 9µB5Bext
3 2 5/2 −1/2 6/5 − 3µB5Bext
3 2 5/2 +1/2 6/5 + 3µB5Bext
3 2 5/2 +3/2 18/5 + 9µB5Bext
3 2 5/2 +5/2 6 +3µB Bext

There are total 10 transitions which are possible from the n = 3 levels to the n = 1 levels.
They are,

3p1/2 −→ 1s1/2 4 transitions

3p3/2 −→ 1s1/2 6 transitions.

No transitions are possible between 3s and 1s, 3d and 1s.

(b) Strong Field
The Table of energies, for the hydrogen atom when placed in a strong magnetic field, is
shown in Figure 2. The shift in energy for the strong field is,

∆Estrong = µB (ml + 2ms ) Bext .

This perturbation is applied before the spin-orbit interaction. The Fine-structure correction

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Atomic, molecular and laser physics

is,
( )
En2 ml ms − 4l(l + 1)
∆Ef s = + 3 , and is applicable for l > 0 only.
2mc2 l(l + 1)(l + 1/2)

n l ml ms ml + 2ms ∆EZeeman ∆Efs

3 0 0 −1/2 −1 −µB Bext 0
3 0 0 +1/2 +1 +µB Bext 0
2
3 1 −1 −1/2 −2 −2µB Bext En
4mc2
2
3 1 0 −1/2 −1 −µB Bext En
6mc2
2
3 1 1 −1/2 0 0 En
12mc2
2
3 1 −1 +1/2 0 0 En
12mc2
En2
3 1 0 +1/2 +1 +µB Bext 6mc2
En2
3 1 1 +1/2 +2 +2µB Bext 4mc2
2
3 2 −2 −1/2 −3 −3µB Bext 11En
15mc2
2
3 2 −1 −1/2 0 0 43En
60mc2
2
3 2 0 −1/2 −1 −µB Bext 7En
10mc2
2
3 2 1 −1/2 −2 −2µB Bext 41En
60mc2
2
3 2 2 −1/2 +1 +µB Bext 2En
3mc2
2
3 2 −2 +1/2 −1 −µB Bext 11En
15mc2
2
3 2 −1 +1/2 +2 +2µB Bext 41En
60mc2
7En2
3 2 0 +1/2 +1 +µB Bext 10mc2
43En 2
3 2 1 +1/2 0 0 60mc2
11En 2
3 2 2 +1/2 +3 +3µB Bext 15mc2

There is total 4 transitions which are possible from n = 3 level to n = 1 energy level. They
are,

3p −→ 1s 4 transitions.

The selection rules used are ∆l = ±1, ∆ml = 0, ±1, (transition from 0 to 0 is not allowed)
and ∆ms = 0.
Solution 2.
The distinct spectral lines resulting from 2p1/2 −→ 1s1/2 are shown in the Figure 3. When
placed inside a weak magnetic field, the correction to the energy relative to the fine-structure

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Atomic, molecular and laser physics

(2,1/2) = 3µΒΒext
(-1,1/2) = 2 µΒΒext
(0,1/2) (2,-1/2) = µΒΒ
ext
(1,1/2) (-1,-1/2) = 0
3d (0,-1/2) (-2,1/2)
=−µΒΒext
(1,-1/2) = −2 µΒΒext
(-2,-1/2) = −3µΒΒext

(1,1/2) = 2 µΒΒext
(0,1/2) = µΒΒ ext (0,1/2) = µΒΒ ext
(-1,1/2) (1,-1/2) =0
3s (0,-1/2) = −µΒΒext
3p (0,-1/2)
=−µΒΒext
(-1,-1/2)
= −2 µΒΒext

(1,1/2) = 2 µΒΒext
(0,1/2) = µΒΒext (0,1/2) = µΒΒ ext
(-1,1/2) (1,-1/2) =0
2s (0,-1/2) = −µΒΒext
2p (0,-1/2) =−µΒΒext
(-1,-1/2) = −2 µΒΒext

(0,1/2) = µΒΒ ext

1s (0,-1/2)
Ε=−µΒΒext

FIG. 2: Splitting of hydrogen’s n = 3 energy levels when placed in a strong magnetic field. Above
each state, the values of ml , ms are explicitly specified. These lines further split up into different
states (called fine-structure splitting), which are not shown in this diagram.

levels is,

∆E = gJ µB mj Bext

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Atomic, molecular and laser physics

where,

j(j + 1) + 3/4 − l(l + 1)

gJ = 1 +
2j(j + 1)
mj = j, j − 1, . . . , −j + 1, −j, and
e~
µB = ·
2me

(1/2,1/2) Ε=µΒΒext
3
2p1/2
(1/2, -1/2) Ε=−µΒΒext
3

υ Ε=µΒΒext
(1/2,1/2)

1s1/2
(1/2, -1/2) Ε=−µΒΒ ext

FIG. 3: The distinct spectral lines resulting from 2p1/2 −→ 1s1/2 transitions. For each energy
level, values of j, mj are explicitly specified in brackets. The ∆E ′ s mentioned are with respect
to the degenerate levels.

The wavenumbers corresponding to the 2p1/2 −→ 1s1/2 transitions are,

2p(1/2,−1/2) −→ 1s(1/2,−1/2) ν̄
2µB Bext
2p(1/2,+1/2) −→ 1s(1/2,−1/2) ν̄ +
3hc
2µB Bext
2p(1/2,−1/2) −→ 1s(1/2,+1/2) ν̄ −
hc
2µB Bext 2µB Bext
2p(1/2,+1/2) −→ 1s(1/2,+1/2) ν̄ − + ·
hc 3hc

The selection rules used are ∆j = 0, ±1, ∆mj = 0, ±1, ∆l = ±1.

Similarly, for the transition 2p3/2 −→ 1s1/2 , the distinct spectral lines are shown in Figure 4.

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Atomic, molecular and laser physics

(3/2,3/2) Ε= 2µΒΒext
(3/2,1/2) Ε= 2µΒΒext
3
2p3/2
(3/2, -1/2) Ε=−2µΒΒext
3
(3/2, -3/2) Ε=−2µΒΒext

(1/2,1/2)
υ Ε= µΒΒext

1s1/2
(1/2, -1/2) Ε=−µΒΒext

FIG. 4: The distinct spectral lines resulting from 2p3/2 −→ 1s1/2 transitions. For each energy level,
values of j, mj are specified explicitly in brackets. Once again, the ∆E ′ s are with respect to
the degenerate levels.

The wavenumbers corresponding to these transitions are,

2p(3/2,−3/2) −→ 1s(1/2,−1/2) ν̄
4µB Bext
2p(3/2,−1/2) −→ 1s(1/2,−1/2) ν̄ +
(3hc )
4µB Bext
2p(3/2,+1/2) −→ 1s(1/2,−1/2) ν̄ + 2
3hc
2µB Bext 4µB Bext
2p(3/2,−1/2) −→ 1s(1/2,+1/2) ν̄ − +
hc (3hc )
2µB Bext 4µB Bext
2p(3/2,+1/2) −→ 1s(1/2,+1/2) ν̄ − +2
hc 3hc
( )
2µB Bext 4µB Bext
2p(3/2,+3/2) −→ 1s(1/2,1/2) ν̄ − +3 ·
hc 3hc

Solution 3.
To draw the energy level diagram for the hydrogen atom (I = 1/2), first we need to know
the values of F for different j. The total quantum number F is given by F = j + I, j + I −
1 , . . . |j − I|. Therefore,
For n=1

l = 0, I = 1/2, j = 1/2, F = 0, 1.

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Atomic, molecular and laser physics

For n=2

l = 0, 1, I = 1/2, j = 1/2, F = 0, 1
j = 3/2, F = 1, 2.

For n=3

l = 0, 1, 2, I = 1/2, j = 1/2, F = 0, 1
j = 3/2, F = 1, 2
j = 5/2, F = 1, 2, 3.

All these energy levels are shown in Figure 5.

Similarly, for a nucleus with I = 3/2, the total quantum number F for different j ′ s is,
For n=1

l = 0, I = 3/2, j = 1/2, F = 1, 2.

For n=2

l = 0, 1, I = 3/2, j = 1/2, F = 1, 2

j = 3/2, F = 0, 1, 2, 3.

For n=3

l = 0, 1, 2, I = 3/2, j = 1/2, F = 1, 2

j = 3/2, F = 0, 1, 2, 3
j = 5/2, F = 1, 2, 3, 4.

All these energy states are shown in Figure 6.

Solution 4.
For the 3d level of hydrogen,

n = 3, l = 2, j = 3/2, mj = −3/2, −1/2, 1/2, 3/2, and

j = 5/2, mj = −5/2, −3/2, −1/2, 1/2, 3/2, 5/2.

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Atomic, molecular and laser physics

Hydrogen (I=1/2)

n=3
F=3
F=2
3d5/2 3d5/2
F=1

F=2 F=2
3p3./2 3d3/2 3p3./2 3d3/2
F=1 F=1

F=1 F=1
3s1/2 3p1/2 3s1/2 3p1/2
n=2 F=0 F=0

F=2
2p3/2 2p3/2
F=1

F=1 F=1
2s1/2 2p1/2 2s1/2 2p1/2
F=0 F=0
n=1
F=1
1s1/2 1s1/2
F=0
Spin-Orbit Coupling Hyperfine Interaction

FIG. 5: Grotrion diagram for hydrogen (I = 1/2).

⃗ is forming with z-axis is when

The minimum angle that the total angular momentum (J)

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Nucleus with I=3/2

n=3 F=4
F=3
3d5/2 3d5/2 F=2
F=1

F=3 F=3
F=2 F=2
3p3./2 3d3/2 3p3./2 F=1 3d3/2 F=1
F=0 F=0

F=2 F=2
3s1/2 3p1/2 3s1/2 3p1/2
n=2 F=1 F=1
F=3
F=2
2p3/2 2p3/2 F=1
F=0

F=2 F=2
2s1/2 2p1/2 2s1/2 2p1/2
F=1 F=1
n=1
F=2
1s1/2 1s1/2
F=1
Spin-Orbit Coupling Hyperfine Interaction

FIG. 6: Grotrion diagram for a nucleus with I = 3/2.

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Atomic, molecular and laser physics

mj has the maximum value (mj = 5/2),

( )
−1 mj ~
θ = cos √
j(j + 1) ~
( )
−1 5/2~
= cos √
5/2(5/2 + 1) ~
( )
= cos−1 0.845
= 32.3 ◦ .

Solution 5. The difference in energy between the two kinds of photon is,
( )
1 1
∆E = hc −
λ1 λ2
( )
1 1
= hc −
589.0 × 10−9 m 589.6 × 10−9 m
= 3.43 × 10−22 J

= 2.15 × 10−3 eV.

The origin of this energy lies in the interaction of magnetic moments, arising from the
angular momentum S ⃗ and L
⃗ of an electron.

∆E ≈ −2µ⃗s .B
⃗ int
e ⃗
µ⃗s = −gs S, where gs = 2 .
2me
Suppose that m ⃗ int are parallel (say both pointing along z−direction), then,
⃗ s and B
e e
|µ⃗s | = Sz = ms ~
me me
e
∆E = −2 ms ~ Bint
me
e
3.43 × 10−22 J = − ~Bint
me
3.43 × 10−22 J × 9.11 × 10−31 kg
|Bint | ≈
1.6 × 10−19 J × 1.054 × 10−34 J.sec
≈ 18 T.

Solution 6.
From the electron’s point of view, it is the proton that circles around and this orbital motion
creates a magnetic field at the center. Therefore, the spin angular momentum (S) ⃗ and the
⃗ associated with an elecron are now interacting through their
orbital angular momentum (L)
magnetic moments.

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Atomic, molecular and laser physics

The Hamiltonian for spin-orbit coupling, including the Thomas precession effect (which adds
a factor of 1/2), is given by,
e2 ⃗ L.
⃗
Hs−o = S.
8πm2 c2 εr3
⃗ nor L
In the presence of spin-orbit coupling, neither S ⃗ is separately conserved; the conserved

quantity is the total angular momentum, J⃗ = L ⃗ + S.

⃗

J⃗2 = L
⃗2 + S
⃗ 2 + 2L.
⃗ S⃗
⃗L
S. ⃗ = 1/2(J⃗2 − L
⃗2 − S
⃗ 2)

Therefore, the change in energy is,

∆Es−o = ⟨n, l, j, mj |Hs−o |n, l, j, mj ⟩

e2 ⃗L
S. ⃗
= ⟨n, l, j, m j | |n, l, j, mj ⟩
8πm2 c2 ε r3
e2 (J⃗2 − L⃗2 − S ⃗ 2)
= ⟨n, l, j, m j | |n, l, j, mj ⟩
8πm2 c2 ε 2r3
e2 1 1[ ]
= ⟨ ⟩ j(j + 1)~ 2
− l(l + 1)~ 2
− 3/4~ 2
,
8πm2 c2 ε 2r3 2
where,

J 2 |ϕ⟩ = j(j + 1)~2 |ϕ⟩

L2 |ϕ⟩ = l(l + 1)~2 |ϕ⟩

S 2 |ϕ⟩ = s(s + 1)~2 |ϕ⟩

S 2 |ϕ⟩ = 3/4~2 |ϕ⟩ (s = 1/2).

Note that the good quantum numbers after including the spin-orbit interaction are n, l, j, mj
instead of n, l, ml , ms .
e2 [ ] 1
∆Es−o = j(j + 1)~ 2
− l(l + 1)~ 2
− 3/4~ 2
⟨ 3 ⟩.
16πm2 c2 ε 2r
Using,
1 1
⟨ 3
⟩=
2r l(l + 1/2)(l + 1)n3 a3

e2 [ ] 1
∆Es−o = 2 2
j(j + 1)~ 2
− l(l + 1)~ 2
− 3/4~ 2
16πm c ε l(l + 1/2)(l + 1)n3 a3
( )
e2 ~ 2 j(j + 1) − l(l + 1) − 3/4
= 2 2 3 3
·
16πm c a εn l(l + 1/2)(l + 1)

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Now using the expression for the Bohr radius,

4πε~
a = ,
me4 ( )
e2 ~ 2 j(j + 1) − l(l + 1) − 3/4
∆Es−o =
16πm2 c2 εn3 l(l + 1/2)(l + 1)
( )
me8 j(j + 1) − l(l + 1) − 3/4
= ·
(64 × 16)π 4 c2 ε4 ~4 n3 l(l + 1/2)(l + 1)
Since the Bohr energies are given by,
( 2 )2
me e
En(0) =
2 4πε~
and from the definition of fine structure constant,
e2
α = ,
4πε~c
We have
( 2 )2 ( 2 )2 ( )
me e e 1 j(j + 1) − l(l + 1) − 3/4
∆Es−o =
4n2 4πε~ 4πε~c n l(l + 1/2)(l + 1)
(0) 2 ( )
En α j(j + 1) − l(l + 1) − 3/4
= ,
2n l(l + 1/2)(l + 1)
indicating that the shift in energy due to the spin-orbit interaction is proportional to α2 .
Solution 7.
In the 4f -level of hydrogen, n = 4, l = 3, s = 1/2. The total angular momentum j can have
value in the range j = l + s, l + s − 1, . . . |l − s|. Therefore, j = 5/2, 7/2.
As a result, the 4f -level of hydrogen splits up into two levels due to the spin-orbit inter-
actions. The energies corresponding to these fine-structure levels (relative to unperturbed
levels) is,
( )
E1 α 2 3 2n
∆Ef s = − ·
2n4 2 j + 1/2
7E1 α2
For 4f5/2 , ∆Ef s = − ,
3072
E1 α 2
and for 4f7/2 , ∆Ef s = − ·
1024
When we place the 4f levels in a weak magnetic field, they split up further into 2j + 1
states. Therefore, the level with j = 5/2 is 2j + 1 = 6-fold degenerate and the level j = 7/2
is 2j + 1 = 8-fold degenerate; 4f5/2 splits into 6 states and 4f7/2 splits into 8 states, as shown
in Figure 7.

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(j,mj)
(7/2,7/2)
(7/2,5/2)
(7/2,3/2)
(7/2,1/2)
4f7/2
(7/2,-1/2)
(7/2,-3/2)
(7/2,-5/2)
4f (7/2,-7/2)

(5/2,5/2)
(5/2,3/2)
(5/2,1/2)
(5/2,-1/2)
4f5/2
(5/2,-3/2)
(5/2,-5/2)
(j,mj)
Unperturbed Fine-structure Inside weak
levels (including magnetic field
spin-orbit
interaction)

FIG. 7: The fine-structure and Zeeman splitting of hydrogen 4f levels.

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