Solution Set # 7 Zeeman Effect and Hyperfine Interaction

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 µBmjBext

where,

j(j + 1) + 3/4 − l(l + 1) g = 1 + J 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) ext = 9/5 µΒΒext (5/2,1/2) = 3/5 µΒΒext (5/2,-1/2) 3d 5/2 = − 3/5 µΒΒext (5/2,-3/2) = − 9/5 µΒΒext (5/2,-5/2) = 3µΒΒext (3/2,3/2) (3/2,3/2) = 2µΒΒext = 6/5 µΒΒext (3/2,1/2) (3/2,1/2) = 2/3 µΒΒext = 2/5 µΒΒext

(3/2,-1/2) (3/2,-1/2) 2/5 Β ext 3p 3/2 = − 2/3 µΒΒext 3d 3/2 = − µ Β (3/2,-3/2) (3/2,-3/2) = − 2µΒΒext = − 6/5 µΒΒext (1/2,1/2) = µ ΒΒext 1/3 Β ext (1/2,1/2) = µ Β

3s 1/2 (1/2,-1/2) 3p 1/2 (1/2,-1/2) =−µ ΒΒext = − 1/3 µΒΒext

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

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

2s 1/2 (1/2,-1/2) 2p 1/2 (1/2,-1/2) = −µ ΒΒext = − 1/3 µΒΒext

(1/2,1/2) = µ ΒΒext

1s 1/2 (1/2,-1/2) Ε=−µ ΒΒext

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 −µBBext

3 0 1/2 +1/2 2 +µBBext

− − µB Bext 3 1 1/2 1/2 2/3 3

µB Bext 3 1 1/2 +1/2 2/3 + 3

3 1 3/2 −3/2 4 −2µBBext

− − 2µB Bext 3 1 3/2 1/2 4/3 3

2µB Bext 3 1 3/2 +1/2 4/3 + 3

3 1 3/2 +3/2 4 +2µBBext

− − 6µB Bext 3 2 3/2 3/2 12/5 5 − − 2µB Bext 3 2 3/2 1/2 4/5 5

2µB Bext 3 2 3/2 +1/2 4/5 + 5

6µB Bext 3 2 3/2 +3/2 12/5 + 5

3 2 5/2 −5/2 6 −3µBBext

− − 9µB Bext 3 2 5/2 3/2 18/5 5 − − 3µB Bext 3 2 5/2 1/2 6/5 5

3µB Bext 3 2 5/2 +1/2 6/5 + 5

9µB Bext 3 2 5/2 +3/2 18/5 + 5

3 2 5/2 +5/2 6 +3µBBext

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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is, ( ) E2 m m − 4l(l + 1) ∆E = n l s + 3 , and is applicable for l > 0 only. fs 2mc2 l(l + 1)(l + 1/2)

n l ml ms ml + 2ms ∆EZeeman ∆Efs

3 0 0 −1/2 −1 −µBBext 0

3 0 0 +1/2 +1 +µBBext 0 2 − − − − En 3 1 1 1/2 2 2µBBext 4mc2 2 − − − En 3 1 0 1/2 1 µBBext 6mc2 2 − En 3 1 1 1/2 0 0 12mc2 2 − En 3 1 1 +1/2 0 0 12mc2 2 En 3 1 0 +1/2 +1 +µBBext 6mc2 2 En 3 1 1 +1/2 +2 +2µBBext 4mc2 2 − − − − 11En 3 2 2 1/2 3 3µBBext 15mc2 2 − − 43En 3 2 1 1/2 0 0 60mc2 2 − − − 7En 3 2 0 1/2 1 µBBext 10mc2 2 − − − 41En 3 2 1 1/2 2 2µBBext 60mc2 2 − 2En 3 2 2 1/2 +1 +µBBext 3mc2 2 − − − 11En 3 2 2 +1/2 1 µBBext 15mc2 2 − 41En 3 2 1 +1/2 +2 +2µBBext 60mc2 2 7En 3 2 0 +1/2 +1 +µBBext 10mc2 2 43En 3 2 1 +1/2 0 0 60mc2 2 11En 3 2 2 +1/2 +3 +3µBBext 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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(2,1/2) = 3µΒΒext (-1,1/2) = 2 µΒΒext (0,1/2) (2,-1/2) = µ ΒΒext (1,1/2) (-1,-1/2) = 0 3d (0,-1/2) (-2,1/2) =−µ ΒΒext (1,-1/2) = − 2 µΒΒext (-2,-1/2) = − 3µΒΒext (1,1/2) (0,1/2) = 2 µΒΒext = µ ΒΒext (0,1/2) = µ ΒΒext (-1,1/2) (1,-1/2) = 0 3s (0,-1/2) 3p (0,-1/2) = −µ ΒΒext (-1,-1/2) =−µ ΒΒext = − 2 µΒΒext

(1,1/2) = 2 µΒΒext (0,1/2) = µ ΒΒext (0,1/2) (-1,1/2) (1,-1/2) = µ ΒΒext = 0 (0,-1/2) 2s (0,-1/2) 2p =−µ ΒΒext = −µ ΒΒext (-1,-1/2) = − 2 µΒΒext

(0,1/2) = µ ΒΒext 1s (0,-1/2) Ε=−µ ΒΒext

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 µBmjBext

5 Atomic, molecular and laser physics where,

j(j + 1) + 3/4 − l(l + 1) g = 1 + J 2j(j + 1)

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

(1/2,1/2) Ε=µ ΒΒext 3

2p 1/2 (1/2, - 1/2) Ε=−µ ΒΒext 3

υ (1/2,1/2) Ε=µ ΒΒext

1s 1/2 (1/2, - 1/2) Ε=−µ ΒΒext

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) ν¯ 2µBBext 2p −→ 1s − ν¯ + (1/2,+1/2) (1/2, 1/2) 3hc 2µBBext 2p − −→ 1s ν¯ − (1/2, 1/2) (1/2,+1/2) hc 2µ B 2µ B 2p −→ 1s ν¯ − B ext + B ext · (1/2,+1/2) (1/2,+1/2) 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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(3/2,3/2) Ε= 2µ ΒΒext (3/2,1/2) Ε= 2µ ΒΒext 3 2p 3/2 (3/2, - 1/2) Ε=−2µ ΒΒext 3 (3/2, - 3/2) Ε=−2µ ΒΒext

(1/2,1/2) υ Ε= µ ΒΒext

1s 1/2 (1/2, - 1/2) Ε=−µ ΒΒext

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) ν¯ 4µBBext 2p(3/2,−1/2) −→ 1s(1/2,−1/2) ν¯ + (3hc ) 4µBBext 2p −→ 1s − ν¯ + 2 (3/2,+1/2) (1/2, 1/2) 3hc

2µBBext 4µBBext 2p(3/2,−1/2) −→ 1s(1/2,+1/2) ν¯ − + hc (3hc ) 2µBBext 4µBBext 2p(3/2,+1/2) −→ 1s(1/2,+1/2) ν¯ − + 2 hc ( 3hc ) 2µ B 4µ B 2p −→ 1s ν¯ − B ext + 3 B ext · (3/2,+3/2) (1/2,1/2) 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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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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Hydrogen (I=1/2)

n=3 F=3 F=2 3d 5/2 3d 5/2 F=1

F=2 F=2

3p 3./2 3d 3/2 3p 3./2 3d 3/2 F=1 F=1

F=1 F=1

3s 1/2 3p 1/2 3s 1/2 3p 1/2 n=2 F=0 F=0

F=2

2p 3/2 2p 3/2 F=1

F=1 F=1 2s 1/2 2p 1/2 2s 1/2 2p 1/2 F=0 F=0 n=1 F=1

1s 1/2 1s 1/2 F=0 Spin-Orbit Coupling Hyperfine Interaction

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

The minimum angle that the total angular momentum (J⃗) is forming with z-axis is when

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Nucleus with I=3/2

n=3 F=4 F=3 3d 5/2 3d 5/2 F=2 F=1

F=3 F=3 F=2 F=2 3p 3./2 3d 3/2 3p 3./2 F=1 3d 3/2 F=1 F=0 F=0

F=2 F=2

3s 1/2 3p 1/2 3s 1/2 3p 1/2 n=2 F=1 F=1 F=3 F=2 2p 3/2 2p 3/2 F=1 F=0

F=2 F=2

2s 1/2 2p 1/2 2s 1/2 2p 1/2 F=1 F=1 n=1 F=2

1s 1/2 1s 1/2 F=1 Spin-Orbit Coupling Hyperfine Interaction

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

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mj has the maximum value (mj = 5/2), ( ) m ~ θ = cos−1 √ j ~ ( j(j + 1) ) 5/2~ = cos−1 √ ~ ( 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 S⃗ and L⃗ of an electron.

⃗ ∆E ≈ −2 ⃗µs.Bint e ⃗ ⃗µs = −gs S, where gs = 2 . 2me ⃗ Suppose that ⃗ms and Bint are parallel (say both pointing along z−direction), then, e e | ⃗µs| = Sz = ms~ me me e ∆E = −2 ms~ Bint me −22 e 3.43 × 10 J = − ~Bint me 3.43 × 10−22 J × 9.11 × 10−31 kg |B | ≈ int 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 (S⃗) and the orbital angular momentum (L⃗ ) associated with an elecron are now interacting through their magnetic moments.

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The Hamiltonian for spin-orbit coupling, including the Thomas precession eﬀect (which adds a factor of 1/2), is given by, e2 H − = S.⃗ L.⃗ s o 8πm2c2εr3 In the presence of spin-orbit coupling, neither S⃗ nor L⃗ is separately conserved; the conserved quantity is the total angular momentum, J⃗ = L⃗ + S⃗.

J⃗2 = L⃗ 2 + S⃗2 + 2L.⃗ S⃗

S.⃗ L⃗ = 1/2(J⃗2 − L⃗ 2 − S⃗2)

Therefore, the change in energy is,

∆Es−o = ⟨n, l, j, mj|Hs−o|n, l, j, mj⟩ e2 S.⃗ L⃗ = ⟨n, l, j, m | |n, l, j, m ⟩ 8πm2c2ε j r3 j e2 (J⃗2 − L⃗ 2 − S⃗2) = ⟨n, l, j, m | |n, l, j, m ⟩ 8πm2c2ε j 2r3 j e2 1 1[ ] = ⟨ ⟩ j(j + 1)~2 − l(l + 1)~2 − 3/4~2 , 8πm2c2ε 2r3 2 where,

J 2|ϕ⟩ = j(j + 1)~2|ϕ⟩

L2|ϕ⟩ = l(l + 1)~2|ϕ⟩

S2|ϕ⟩ = s(s + 1)~2|ϕ⟩

S2|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.

2 [ ] e 2 2 2 1 ∆E − = j(j + 1)~ − l(l + 1)~ − 3/4~ ⟨ ⟩. s o 16πm2c2ε 2r3 Using, 1 1 ⟨ ⟩ = 2r3 l(l + 1/2)(l + 1)n3a3

2 [ ] e 2 2 2 1 − ~ − ~ − ~ ∆Es o = 2 2 j(j + 1) l(l + 1) 3/4 3 3 16πm c ε ( ) l(l + 1/2)(l + 1)n a e2~2 j(j + 1) − l(l + 1) − 3/4 = · 16πm2c2a3εn3 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 = 2 2 3 16πm c εn l(l(+ 1/2)(l + 1) ) me8 j(j + 1) − l(l + 1) − 3/4 = · (64 × 16)π4c2ε4~4n3 l(l + 1/2)(l + 1) Since the Bohr energies are given by, ( ) m e2 2 E(0) = e n 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 ∆E − = s o 4n2 4πε~ 4πε~c n l(l + 1/2)(l + 1) ( ) E(0)α2 j(j + 1) − l(l + 1) − 3/4 = n , 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, ( ) E α2 3 2n ∆E = 1 − · fs 2n4 2 j + 1/2 7E α2 For 4f , ∆E = − 1 , 5/2 fs 3072 E α2 and for 4f , ∆E = − 1 · 7/2 fs 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,m j) (7/2,7/2) (7/2,5/2) (7/2,3/2) (7/2,1/2) 4f 7/2 (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) 4f 5/2 (5/2,-3/2) (5/2,-5/2) (j,m j)

Unperturbed Fine-structure Inside weak levels (including magnetic field spin-orbit interaction )

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