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Home , Electron magnetic moment, Moment (physics), Orbital motion (quantum), Transition dipole moment

Chapter 19 Magnetism, Angular , and

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 19: Magnetism, , and Spin In 1820 Hans Christian Ørsted discovered that electric current produces a magnetic field that deflects compass needle from magnetic north, establishing first direct connection between fields of electricity and magnetism.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Biot-Savart Law Jean-Baptiste Biot and Félix Savart worked out that magnetic field, B⃗, produced at distance r away from section of wire of length dl carrying steady current  is

휇  d⃗l × ⃗r dB⃗ = 0 Biot-Savart law 4휋 r3

Direction of magnetic field vector is given by “right-hand” rule: if you point thumb of your right hand along direction of current then your fingers will curl in direction of magnetic field.

current

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Microscopic Origins of Magnetism Shortly after Biot and Savart, Ampére suggested that magnetism in matter arises from a multitude of ring currents circulating at atomic and molecular scale.

André-Marie Ampére 1775 - 1836

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Magnetic moment from current loop Current flowing in flat loop of wire with area A will generate magnetic field magnetic which, at distance much larger than radius, r, appears identical to field dipole produced by point with strength of radius 휇 = | ⃗휇| =  ⋅ A

current Example What is magnetic dipole moment induced by e− in circular orbit of radius r with linear v? − 휋 Solution: For e with linear velocity of v the for one orbit is torbit = 2 r∕v. Taking current as charge, −qe, passing point per unit time we have q v  = −q ∕t = − e e orbit 2휋r Taking area of current loop as A = 휋r2 we obtain magnetic dipole moment of ( ) −q v q 휇 = A = e (휋r2) = − e r v 2휋r 2 P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Orbital Magnetic Dipole q Recalling 휇 = − e r v, for electron with angular momentum, rearrange L⃗ = m⃗r × v⃗, to get 2 ⃗ q ⃗ L ⃗ ⃗ ⃗휇 e L 훾 ⃗ 훾 ≡ = r × v and then obtain = − = − LL where L qe∕(2me) m 2 me 훾 L is called orbital : ratio of to orbital angular momentum

magnetic Angular momentum vector is antiparallel to magnetic dipole dipole vector due to negative electron charge. Don’t forget : current flow is defined in opposite direction of electron flow. radius ⃗휇 훾 ⃗ Despite simple derivation, = − LL is general and holds for current non-circular orbits as long as angular momentum is conserved.

Angular ⃗휇 훾 ⃗ = − LL is independent of size, period, and even shape of momentum orbit.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Electron Orbital Magnetic Dipole From e− orbital the magnetic dipole moment operator is q 휇 ⃗̂휇 e ⃗̂ B ⃗̂ 퓁 = − L = − ℏ L 2me 휇 B is defined as (an atomic unit for dipole moments) and given by q ℏ 휇 ≡ e . −24 B = 9 274009992054043 × 10 J/T 2me √ 휇 휇 퓁 퓁 Magnitude of dipole moment is 퓁 = B ( + 1) Operator for z component of electron orbital magnetic dipole moment is 휇 ̂휇 B ̂ z = − ℏ Lz

For z (or x or y) components only observed with quantized values of 휇 휇 휇 B ⟨̂ ⟩ B ℏ 휇 z = − ℏ Lz = − ℏ (m퓁 ) = − Bm퓁

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical magnetic dipole in a uniform magnetic field of classical magnetic dipole in uniform magnetic field is

x is N ⃗ ⃗ V = − ⃗휇 ⋅ B = −| ⃗휇||B| cos 휃 z S Note: 휃 = 180◦ is an unstable equilibrium point, i.e., potential is a maximum at 휃 = 180◦. Take as K = 1 I휔2, then total energy is 2 1 1 E = I휔2 − ⃗휇 ⋅ B⃗ = I휔2 − | ⃗휇||B⃗| cos 휃 2 2 With the magnetic dipole and magnetic field vectors in the x-z plane, magnetic dipole experiences about y given by [ ] ⃗휏 ⃗휇 ⃗ | ⃗휇||⃗| 휃 ⃗ = × B = B sin ey In absence of total energy remains constant and compass needle oscillates about direction of magnetic field.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Magnetic dipole with angular momentum

Spinning top precesses in gravitational field. Magnetic dipole with angular momentum precesses in magnetic field

Precession Direction

N S

⃗ ⃗ dJ ⃗휏 ⃗ ⃗ dJ = = R × Mg = ⃗휏 = ⃗휇 × B⃗ dt dt

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin e−’s magnetic dipole from orbital motion in uniform magnetic field Zeeman interaction 휇 ⃗ For e− with orbital motion substitute ⃗̂휇 B ̂ into ⃗휇 ⋅ ⃗ and obtain 퓁 = − ℏ L V = − B 휇 ̂ B ⃗̂ ⋅ ⃗ V = − ℏ L B

Coupling of e−’s magnetic dipole moment to external magnetic field is called Zeeman interaction. ⃗ ⃗ Taking magnetic field as along z axis, B = Bzez, expression simplifies to 휇 ̂ B ̂ V = − ℏ LzBz

Zeeman interaction lifts degeneracy of m퓁 levels of electron in H-.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical magnetic dipole with angular momentum in uniform magnetic field Place classical magnetic dipole with anti-parallel angular momentum in magnetic field and magnetic moment experiences torque ⃗휏 ⃗휇 ⃗ 훾 ⃗ ⃗ 훾 ⃗ ⃗ = × B = − LL × B = LB × L Torque causes change in angular momentum. Newton’s 2nd law says

dL⃗ = ⃗휏 = 훾 B⃗ × L⃗ dt L

⃗ ⃗ Taking B = Bzez, and defining 훾 Ω0 = − LBz we write in determinant form | | | e⃗ e⃗ e⃗ | ⃗ | x y z | dL | | ⃗ ⃗ = | 0 0 −Ω0 | = LyΩ0ex − LxΩ0ey dt | | | Lx Ly Lz |

Determinant gives 3 coupled differential equations...

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical magnetic dipole with angular momentum in uniform magnetic field

dL dLy dL x = Ω L , = −Ω L , z = 0 dt 0 y dt 0 x dt Coupled differential eqs. involving Lx and Ly can be decoupled by defining

L± = Lx ± iLy Then we find

dL dL dLy + = x + i = Ω L − iΩ L = −iΩ (L + iL ) = −iΩ L dt dt dt 0 y 0 x 0 x y 0 +

dL dL dLy − = x − i = Ω L + iΩ L = iΩ (L − iL ) = iΩ L dt dt dt 0 y 0 x 0 x y 0 − or dL± = ∓iΩ L whose solution is L (t) = L (0)e∓iΩ0t dt 0 ± ± ± ⃗휇 훾 ⃗ Converting L± back to Cartesian and taking = − LL we find ... P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical magnetic dipole with angular momentum in uniform magnetic field

휇 휇 휇 x(t) = x(0) cos Ω0t + y(0) sin Ω0t 휇 휇 휇 y(t) = y(0) cos Ω0t − x(0) sin Ω0t 휇 휇 z(t) = z(0)

These equations for magnetic dipole vector components describe precession motion of vector around magnetic field direction at of Ω0. When magnetic dipole vector has angular momentum torque does not cause dipole to align with B⃗, but rather causes it to precess about direction of B⃗.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical Magnetic dipole in a Non-Uniform

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical Magnetic dipole in a Non-Uniform Magnetic Field

If magnetic dipole is in non-uniform magnetic field, B⃗(⃗r), then acting on 2 poles will not cancel and dipole starts to translate along magnetic field gradient direction. N In this case net is ( ) ⃗ ⃗ ⃗ ⃗휇 ⋅ ⃗ ⃗

Fnet = F+ + F− = ∇ B(r) S

Dipole oriented in magnetic field gradient is pulled up.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin

Classical Magnetic dipole in a Non-Uniform Magnetic Field S

Differently oriented dipole in same magnetic

field gradient is pulled down. N

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical Magnetic dipole in a Non-Uniform Magnetic Field

When dipole is perpendicular to lines of force

S there is no translation of dipole. N

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin

Classical Magnetic dipole in a Non-Uniform Magnetic Field

S N

S

N

N S

Note that magnetic field lines also include constant magnetic field component. Also expect this magnetic field to exert torque on all three causing them to re-orient their north poles upwards. All three dipole orientations would eventually be pulled up in this magnetic field.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Magnetic dipoles with angular momentum

On the other hand, if the magnetic dipoles also have angular momentum...

S N

S

N

N S

...then torque from constant magnetic field component causes ⃗휇 to precess about constant magnetic field direction instead of reorienting.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical Magnetic dipole in a Non-Uniform Magnetic Field Now with Angular Momentum Take example of magnetic field given by ⃗ ⃗ ⃗ ⃗ B(r) = −gxex + (B0 + gz)ez g is gradient strength. Maxwell’s equations require ∇ ⋅ B⃗ = 0, so need gradient along x to satisfy this constraint. Thus we find ( ) ( ) ( ) ⃗ ⃗휇 ⋅ ⃗ ⃗ ⃗휇 ⋅ ⃗ ⃗ 휇 ⃗ 휇 ⃗ Fnet = −∇V = ∇ B(r) = ∇ (−gxex + (B0 + gz)ez = ∇ −gx xex + (B0 + gz) zez 휇 ⃗ 휇 ⃗ = −g xex + g zez Unlike compass needle, we want to consider magnetic dipole with angular momentum. Thus, ⃗휇 precesses about magnetic field direction while translating. For this precession motion we use 휇 휇 휇 x(t) = x(0) cos Ω0t + y(0) sin Ω0t 휇 휇 휇 y(t) = y(0) cos Ω0t − x(0) sin Ω0t 휇 휇 z(t) = z(0)

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Classical Magnetic dipole in a Non-Uniform Magnetic Field Now with Angular Momentum Using ⃗ 휇 ⃗ 휇 ⃗ Fnet = −g x(t)ex + g z(t)ez ≫ , ⃗휇 assume B0 gx gz and use (t) for precessing magnetic dipole moment to obtain ( ) ⃗ 휇 휇 ⃗ 휇 ⃗ Fnet(t) = −g x(0) cos Ω0t + y(0) sin Ω0t ex + g z(0) ez ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏟⏟

Fx(t) Fz

Time independent Fz component of force will push dipole in z direction—which way depends on initial orientation of dipole.

Time dependent Fx(t) component of force will push dipole back and forth about its original position with a time averaged of zero along x. Bottom line: passing magnetic dipole through magnetic field gradient along z 휇 provides means of measuring z(0). Basis for seminal experiment in 1922 by Otto Stern and Walther Gerlach.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin The Stern-Gerlach Experiment - First Some Background In 1896 Pieter Zeeman discovered clue about magnetic properties of yet unexplained constituents of atom when he reported broadening of yellow D-lines of sodium in flame when held inside strong magnetic field. Became known as Zeeman effect. Bohr’s 1913 theory of atom predicted that e− orbital angular momentum would be quantized. In 1916 Arnold Sommerfeld refined Bohr’s theory to allow for elliptical orbits described by 3 quantum numbers: n, k, and m, where ▶ n = 1, 2, 3, …, called the principal , ▶ k = 1, … , n, called the ▶ m = −k, −k + 1, … , m ≠ 0, … , k − 1, k, called the . m described quantization of z-component of e−’s orbital angular momentum. Refinements allowed Sommerfeld to explain Zeeman effect through dependence of orbital magnetic dipole moment on magnetic quantum number.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin The Stern-Gerlach Experiment

In 1922 Otto Stern and Walther Gerlach designed apparatus in attempt to measure this quantization with of neutral Ag .

Since Ag atom had one valence electron it was predicted in Bohr-Sommerfeld theory that ground state for this electron was n = k = 1 and m = ±1.

Beam of neutral Ag atoms emitted from oven containing Ag metal vapor were collimated and sent into vacuum region of non-uniform magnetic field where atoms are pushed up or down in magnetic field gradient depending on sign of z component of atom’s magnetic dipole moment.

Detector was glass plate onto which atoms exiting non-uniform magnetic field were deposited.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Schematic diagram of the Stern Gerlach apparatus.

Measurement result

Classical N prediction

S

Atom beam

Oven Collimator

휇 | ⃗휇| | ⃗휇| Classical prediction is all values of z between − and would be observed. Sommerfeld predicts only z components associated with m = ±1 will be observed. After 1 year of refining apparatus Stern & Gerlach observed beam splitting into 2 distinct lines. P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin “Attached is the experimental proof of directional quantization. We congratulate you on the confirmation of your theory.”

— Postcard from Stern & Gerlach to Neils Bohr, February 8, 1922.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin The Stern-Gerlach Experiment Stern-Gerlach experiment in 1922 was before De Broglie’s hypothesis in 1923 and Schrödinger’s wave equation in 1925. While S-G experiment successfully demonstrated quantization of electron angular momentum, Stern & Gerlach’s interpretation in terms of Sommerfeld model did not allow them to appreciate full significance of their discovery. In 1927 Phipps & Taylor used S-G technique on beam of H atoms. In 1927 Schrödinger’s wave equation and solutions for H atom was known.

Orbital angular momentum of H atom ground state (n = 1, 퓁 = 0, and m퓁 = 0) predicts only 1 − 휇 outcome for z component of e ’s magnetic dipole moment, z = 0.

N

S

Phipps and Taylor experiment gives 2 outcomes. At this point it became clear that Schrödinger’s picture for atom was incomplete.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Electron Spin

Hypothesis (now theory) that explains Stern-Gerlach (and Phipps & Taylor) result is that e− has ⃗휇 ⃗ intrinsic (built-in) magnetic dipole moment, s, arising from intrinsic angular momentum, S, called spin.

Property called spin because if e− is charged ball spinning about its own axis then it would have magnetic dipole moment from this “spinning” motion.

Physical picture of spinning charged ball doesn’t hold water for various reasons.

Main reason is that such a model predicts only integer values for spin angular momentum quantum number.

휇 S-G and P-T experiments give 2 outcomes from z, consistent with s = 1∕2, with ms = ±1∕2.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Electron Spin and the Dirac Wave Equation

In 1929 Paul Dirac (1902-1984) proposed relativistic wave equation for e−, showing it must have intrinsic angular momentum of s=1/2 and magnetic moment of g 휇 ⃗휇 s B ⃗ where . is called electron -factor s = − ℏ S gs = 2 00231930436182 g √ 휇 휇 Magnitude of spin dipole moment is s = gs B s(s + 1) In same theory he also predicted existence of anti-electron (and anti-matter).

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Spin Orbitals Instead of switching to we modify (ad hoc) Schrödinger equation solutions to be product with new degree of freedom

휓(⃗r)휒(S⃗)

휒 ⃗ − 휒 훼 (S) is spin part of e ’s wave function whose stationary states, m , are given symbols for 훽 s ms = +1∕2 and for ms = −1∕2. Spin part is expressed as linear combination ⃗ 휒(S) = c훼훼 + c훽훽

c훼 and c훽 are complex coefficients. Two spin states, 훼 and 훽, are orthogonal,

훼∗훼 휎 훽∗훽 휎 , 훼∗훽 휎 훽∗훼 휎 ∫ d = ∫ d = 1 ∫ d = ∫ d = 0

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Spin Orbitals

휓 ≡ 훼 휓 ≡ 훽 n,0,0, 1 n s , n,0,0,− 1 n s , 2 2 H-atom stationary states modified to 휓 ≡ 훼 휓 ≡ 훽 − n,1,−1, 1 n p−1 , n,1,−1,− 1 n p−1 , include e spin state 2 2 휓 ≡ 훼 휓 ≡ 훽 n,1,0, 1 n p0 , n,1,0,− 1 n p0 , 휓 , 휃, 휙, 휎 휃, 휙 휒 2 2 n,퓁,m퓁,m (r ) = Rn,퓁(r)Y퓁,m퓁 ( ) m 휓 ≡ 훼 휓 ≡ 훽 s s n,1,1, 1 n p1 , n,1,1,− 1 n p1 , 2 2 휓 ≡ 훼 휓 ≡ 훽 and are called spin orbitals. n,2,−2, 1 n d−2 , n,2,−2,− 1 n d−2 , 2 2 휓 ≡ 훼 휓 ≡ 훽 H-atom stationary states are specified by n,2,−1, 1 n d−1 , n,2,−1,− 1 n d−1 , 2 2 4 quantum numbers n, 퓁, m퓁, and m . 휓 ≡ 훼 휓 ≡ 훽 s n,2,0, 1 n d0 , n,2,0,− 1 n d0 , 2 2 휓 ≡ 훼 휓 ≡ 훽 Write H-atom stationary states with n,2,1, 1 n d1 , n,2,1,− 1 n d1 , 2 2 shorthand notation. 휓 ≡ 훼 휓 ≡ 훽 n,2,2, 1 n d2 , n,2,2,− 1 n d2 . 2 2

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Spin Orbitals

Occupancy of H-atom stationary states is often illustrated in form of an orbital diagram. 1 Below is case of n = 1, 퓁 = 0, m퓁 = 0, and m = . s 2

1s 2s 2p 3s 3p 3d

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Electron Spin and Total Magnetic Dipole Moment Operator We complete treatment of e−’s magnetic dipole moment in stating that total magnetic dipole moment for e− includes both orbital and spin contributions, g 휇 g 휇 ⃗̂휇 ⃗̂휇 ⃗̂휇 퓁 B ⃗̂ s B ⃗̂ = 퓁 + s = − ℏ L − ℏ S

g퓁 = 1 and is called orbital g factor. . gs = 2 00231930436182 is called electron g-factor. Total magnetic dipole moment operator for e− is 휇 ( ) ⃗̂휇 B ⃗̂ ⃗̂ = − ℏ g퓁L + gsS

and for total z component of the magnetic dipole moment operator we have 휇 ( ) ̂휇 B ̂ ̂ z = − ℏ g퓁Lz + gsSz

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Nuclear Angular Momentum

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Proton Spin Angular Momentum and Magnetic Dipole Moment In 1924, Pauli suggested existence of spin angular momentum and magnetic dipole of protons In 1933, Otto Stern discovers that proton has intrinsic spin angular momentum and associated intrinsic magnetic dipole moment.

Like electron, proton is I = 1∕2 with mI = ±1∕2. Magnitude of proton magnetic dipole is much smaller than electron’s. q ⃗휇 e ⃗ Like electron, one might expect proton magnetic dipole to be I = I 2mp

Key difference is division by proton , mp, instead of electron mass, me.

Mass ratio predicts proton dipole moment is mp∕me ≈ 1836 smaller than electron Actual proton dipole moment is only 656 times smaller than electron’s. Define nuclear magneton as q ℏ 휇 ≡ e . −27 , N = 5 050783698211084 × 10 J/T 2mp 휇 . 휇 and proton magnetic dipole moment is p = 2 792847351193473 N

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Neutron Spin Angular Momentum and Magnetic Dipole Moment

Soon after Rutherford proposed his model of atom with e− orbiting positively charged nucleus, it was realized that there were inconsistencies between number of positive charges and mass of atom.

In 1920, Rutherford suggests that nucleus also contained neutral particles with mass similar to protons, and called them neutrons.

In 1932, James Chadwick produced 1st convincing evidence of neutron’s existence by bombarding beryllium with alpha particles.

In 1937, Isador Rabi and his group at Columbia University developed the magnetic resonance method to make precise measurements of the magnetic dipole moments of atomic nuclei.

Rabi found that neutron spin quantum number is I = 1∕2 and its magnetic dipole moment is 휇 . 휇 n = −1 913042723136664 N 휇 . 휇 Opposite in sign and smaller than proton magnetic dipole, p = 2 792847351193473 N

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Total angular momentum of atomic nuclei Total angular momentum, ⃗I, of nuclear ground state is determined by number of protons and neutrons in nucleus. General rule from nuclear : ▶ for element with odd atomic number, Z, its most abundant isotope has I > 0, ▶ for element with even atomic number, Z, its most abundant isotope has I = 0

Nuclear magnetic dipole moments are related to I according to

⃗̂휇 훾 ⃗̂, = II 훾 > I is called nuclear gyromagnetic ratio. Only nuclear isotopes with I 0 can have a non-zero nuclear magnetic dipole moment. Rabi’s magnetic resonance technique is the basis for Nuclear Magnetic Resonance (NMR) spectroscopy, Electron Paramagnetic Resonance (EPR) spectroscopy, and Magnetic Resonance Imaging (MRI).

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Nuclear magnetic dipole moments of various (NMR Active) Isotopes 휇 Z N Isotope Mass/(g/mol) Abundance/% I Dipole Moment/( N ) 1 0 1H 1.00782503207 99.985 1/2 2.79284739 2 1 3He 3.01602931914 0.000137 1/2 -2.12762485 3 4 7Li 7.016004548 92.41 3/2 3.2564268 4 5 9Be 9.012182201 100 3/2 -1.1778 5 6 11B 11.009305406 80.2 3/2 2.6886489 6 7 13C 13.00335483778 1.11 1/2 0.7024118 7 7 14N 14.00307400478 99.634 1 0.403761 8 9 17O 16.999131703 0.038 5/2 -1.89379 9 10 19F 18.998403224 100 1/2 2.628868 10 11 21Ne 20.993846684 0.27 3/2 -0.661797 11 12 23Na 22.98976928087 100 3/2 2.21752 12 13 25Mg 24.985836917 10 5/2 -0.85545 13 14 27Al 26.981538627 100 5/2 3.6415069 14 15 29Si 28.9764947 4.683 1/2 -0.55529 15 16 31P 30.973761629 100 1/2 1.1316 16 17 33S 32.971458759 0.75 3/2 0.6438212 17 18 35Cl 34.968852682 75.77 3/2 0.8218743 19 20 39K 38.963706679 93.2581 3/2 0.3914662 20 23 43Ca 42.958766628 0.135 7/2 -1.317643 21 24 45Sc 44.955911909 100 7/2 4.7564866 22 25 47Ti 46.951763088 7.44 5/2 -0.78848 23 28 51V 50.943959507 99.75 7/2 5.14870573 24 29 53Cr 52.940649386 9.501 3/2 -0.47454 25 30 55Mn 54.938045141 100 5/2 3.4532 26 31 57Fe 56.935393969 2.119 1/2 0.09044 27 32 59Co 58.933195048 100 7/2 4.627

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Addition of Angular Momentum

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Addition of Angular Momentum Discovery of electron spin angular momentum means that total electron angular momentum of H-atom becomes J⃗ = L⃗ + S⃗ In classical , this would be calculated trivially,

⎛ Jx ⎞ ⎛ Lx ⎞ ⎛ Sx ⎞ ⃗ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ J = ⎜ Jy ⎟ = ⎜ Ly ⎟ + ⎜ Sy ⎟ ⎝ Jz ⎠ ⎝ Lz ⎠ ⎝ Sz ⎠

But this requires that we know all 3 vector components of L⃗ and S⃗ simultaneously. In QM, we can only know length of each angular momentum vector and one component simultaneously. That is, can only know sum of one component,

Jz = Lz + Sz

which gives a constraint that mj = m퓁 + ms

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Addition of Angular Momentum While QM lacks information needed to calculate length of total angular momentum vector, J⃗, we still have constraints on how big or small the total can be. Largest vector sum is when both vectors point in same direction Smallest vector sum is when both point in opposite directions

When adding angular momentum vectors in QM, we can only constrain total angular momentum as lying between these 2 limits | | ||⃗| |⃗|| ≤ |⃗| ≤ |⃗ ⃗| | L − S | J L + S

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Addition of Angular Momentum | | √ |√ √ | |⃗ ⃗| ≥ |⃗| ≥ ||⃗| |⃗|| ≥ | 퓁 퓁 | L + S J | L − S | translates to j(j + 1) | ( + 1) − s(s + 1)|

constraining j to fall between limits |퓁 − s| ≤ j ≤ 퓁 + s with or half-integral values.

Example What are the total angular momenta resulting from the addition of 퓁 = 1 and s = 1 ? 2 | | | | Solution: j = |1 − 1 | to |1 + 1 | gives j = 1 and 3 . | 2 | | 2 | 2 2 , , , , For each j value there will be 2j + 1 values of mj, that is, mj = −j −j + 1 … j − 1 j Example What are the total angular momenta resulting from the addition of 퓁 = 0 and s = 1 ? 2 | | | | Solution: j = |0 − 1 | to |0 + 1 | gives j = 1 | 2 | | 2 | 2

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Electron Magnetic Moment from Total Angular Momentum Total magnetic dipole moment operator for e− is 휇 ⃗̂휇 B ⃗̂ = −gj ℏ J and for total z component of the magnetic dipole moment operator we have 휇 ̂휇 B ̂ z = −gj ℏ Jz

gj is called Landé g factor:

j(j + 1)(g퓁 + g ) + [퓁(퓁 + 1) − s(s + 1)](g퓁 − g ) g = s s j 2j(j + 1) Magnetic moment associated with electron having given value of j is √ 휇 휇 j = gj B j(j + 1)

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Clebsch-Gordon series expansion 퓁 How do wave functions associated with , m퓁, s, and ms combine to form wave functions associated with j and mj?

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Clebsch-Gordon series expansion

퓁 How do wave functions associated with , m퓁, s, and ms combine to form wave functions associated with j and mj?

They are related through Clebsch-Gordon series expansion,

∑퓁 ∑s 휓 , = C(j, m , 퓁, m퓁, s, m )휓퓁, , , j mj j s m퓁 s ms m퓁=−퓁 ms=−s , , 퓁, , , C(j mj m퓁 s ms) are called Clebsch-Gordon coefficients.

These coefficients are nonzero only when mj = m퓁 + ms.

There are recursive expressions for calculating Clebsch-Gordon coefficients, and tables and calculators are easy to find.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Clebsch-Gordon coefficients

j 1j 2m1m2|j 1j 2JM J − j − j d = ( − 1) 1 2 j 2j 1m2m1|j 2j 1JM √ Square-root sign is to be understood over every coefficient, e.g., for −8∕15 read − 8∕15.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Example 1 1 Construct wave function with j = and mj = − for p-electron in H-atom in terms of wave functions 퓁 2 2 associated with , m퓁, s, and ms. Solution: Addition of 퓁 = 1 and s = 1 leads to 2 possibilities of j = 1 and j = 3 . Focus on j = 1 2 2 2 2 states, and write Clebsch-Gordon series as ∑퓁 ∑s ( ) 1 1 휓 1 = C j = , m , 퓁 = 1, m퓁, s = , m 휓 n,j= ,m j s ,퓁 , , 1 , 2 j 2 2 n =1 m퓁 s= ms m퓁=−퓁 ms=−s 2

Expanding out the m = − 1 case, gives six terms: j 2 ( ) ( ) 1 1 1 1 1 1 1 1 휓 1 1 = C , − , 1, −1, , − n p 훽 + C , − , 1, 0, , − n p 훽 n, ,− 2 2 2 2 −1 2 2 2 2 0 2 2 ( ) ( ) 1 , 1 , , , 1 , 1 훽 1 , 1 , , , 1 , 1 훼 + C − 1 1 − n p1 + C − 1 −1 n p−1 ( 2 2 2 2 ) ( 2 2 2 2) + C 1 , − 1 , 1, 0, 1 , 1 n p 훼 + C 1 , − 1 , 1, 1, 1 , 1 n p 훼 2 2 2 2 0 2 2 2 2 1

1 Only CC coefficients with m = m퓁 + m = − are non-zero. j s 2 P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Example 1 1 Construct wave function with j = and mj = − for p-electron in H-atom in terms of wave functions 퓁 2 2 associated with , m퓁, s, and ms. Solution: We are left with ( ) ( ) 휓 1 , 1 , , , 1 , 1 훽 1 , 1 , , , 1 , 1 훼 n, 1 ,− 1 = C − 1 0 − n p0 + C − 1 −1 n p−1 2 2 2 2 2 2 2 2 2 2 Use 1 × 1 Clebsch-Gordon Coefficient table to find 2 √ 휓 √1 훽 2 훼 n, 1 ,− 1 = n p0 − n p−1 2 2 3 3

Using orbital diagrams, we could represent j = 1 , m = − 1 state as 2 j 2

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Addition of three or more angular momenta

What about addition of three or more angular momenta?

For example, what about 퓁 = 1, s = 1 , and i = 1 , for electron orbital, electron spin, and 2 2 nuclear spin angular momenta, respectively?

We have 3 choices on where to start, and all are valid.

First add 퓁 and s to get j = |퓁 − s| to 퓁 + s, then add j and i to get f = |i − j| to i + j.

or

First add i and s to get j = |i − s| to i + s, then add j and 퓁 to get f = |j − 퓁| to j + 퓁.

or

First add 퓁 and i to get j = |i − 퓁| to i + 퓁, then add j and s to get f = |j − s| to j + s.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Example What are total angular momenta values and number of states that result when 퓁 = 1, s = 1 , 2 and i = 1 are added? 2 Solution: Starting with 퓁 and s, we find that First, for 퓁 = 1 and s = 1 we find j = |퓁 − s| to 퓁 + s gives j = 1 , 3 2 2 2 then, for j = 1 and i = 1 we find that f = |j − i| to j + i gives f = 0, 1 2 2 and for j = 3 and i = 1 we find that f = |j − i| to j + i gives f = 1, 2 2 2 , , , Total angular momentum can be f = 2 1 1 0 each with 2f + 1 values of mf Thus we have

(2 ⋅ 2 + 1) + (2 ⋅ 1 + 1) + (2 ⋅ 1 + 1) + (2 ⋅ 0 + 1) = 12 states

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin of the H-Atom Emission Spectrum

1 Relativistic Correction: p̂ 4 ̂ (1) = − 3 3 8m c0

2 Spin Orbit Interaction: ̂ 휉 ⃗̂ ⋅ ⃗̂ VSO = (r)L S where 휉(r) is the spin-orbit coupling constant.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Relativistic Correction While classical kinetic energy expression is good starting approximation for H-atom, there are slight relativistic effects that can be observed in emission spectra of H-atoms.

p2 mc2 Instead of K = , use K = √ 0 − mc2 2m 2 0 1 − (v∕c0) Derivation in notes shows relativistic correction of e− kinetic energy gives p̂4 ̂ (1) = − 3 3 8m c0 This, in turns, gives a 1st-order relativistic correction to energy eigenvalue of nth state ( ) 2 E(0) { } n 4n E(1) = − 3 r,n 2 퓁 + 1∕2 mec0

(0) where En is zeroth-order energy eigenvalue for nth state. Note: relativistic correction makes energy dependent on 퓁. P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction − − 휇 When e has orbital angular momentum, the e ’s intrinsic spin magnetic dipole, s interacts − with nuclear charge, Zqe, as e orbits around nucleus.

nuclear rest frame electron rest frame

In electron rest frame, we calculate magnetic field generated at electron by orbiting nucleus and calculate spin-orbit interaction as ̃ ⃗휇 ⋅ ⃗ VSO = − s Borb Lengthy derivation in notes gives ... P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction Lengthy derivation in notes gives ̂ 휉 ⃗̂ ⋅ ⃗̂ VSO = (r)L S

휉(r) is spin orbit coupling, given by

1 1 d휙(r) 휉(r) = 2 2 r dr 2mec0 휙(r) is electrostatic potential at electron. Convention is to report spin orbit coupling constant as

2 2 ℏ ∗ ̂ ℏ 휁 ,퓁 = 휓 휉(r)휓 ,퓁, d휏 = ⟨휉(r)⟩ n ∫ n,퓁,m퓁 n m퓁 hc0 hc0 휁 휁 n,퓁 is a wavenumber and hc0 n,퓁 is an energy.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction For H-like atom, we can further calculate 훼2 4 R∞Z 휁 ,퓁 = n n3퓁(퓁 + 1)(퓁 + 1∕2) 훼 is dimensionless constant, defined as the fine structure constant q2 훼 ≡ e . 휋휖 ℏ = 0 007297352566206478 4 0 c0

R∞ is Rydberg constant for H-like atom with nuclear charge Z, 4 meq R = e note: R = R ∕(1 + m ∕m ) ∞ 휖2 3 H ∞ e p 8 0h c0 휁 훼2 . −1 휇 For 2p electron in H-atom, we have 2,1 = R∞∕24 = 0 24 cm or 30.2 eV, which is a small . shift on E2 = −3 4 eV, the 2p energy. Such small perturbation is observable in emission spectrum of H-atoms. 휁 4 VSO plays a bigger role in heavier elements since n,퓁 ∝ Z .

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Perturbation Theory : The Spin Orbit Correction ̂ Let’s calculate 1st-order energy correction from VSO for H-atom. [ ] ( ) 2 ℏ2 1 휕 휕 L̂ 2 Zq ⃗ ⃗ ̂ = ̂ (0) + V̂ where ̂ (0) = − r2 − − e and V̂ = 휉(r)L̂ ⋅ Ŝ SO 2 휕 휕 2 2 휋휖 SO 2me r r r r ℏ 4 0r

In QM, any hermitian operator that commutes with ̂ corresponds to physical quantity that is conserved (i.e., time independent).

⃗ ⃗ ⃗ ⃗ ̂ (0) commutes with both L̂ and Ŝ, i.e., [̂ (0), L̂] = 0 and [̂ (0), Ŝ] = 0, so orbital and spin angular momenta would be separately conserved if ̂ = ̂ (0).

̂ ̂ (0) ̂ ̂ , ⃗̂ ≠ ̂ , ⃗̂ ≠ If spin-orbit coupling is included: = + VSO, then we find [ L] 0 and [ S] 0.

Orbital and spin angular momentum are no longer separately conserved when spin-orbit coupling is present. Still true that total angular momentum, J⃗, is conserved.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Perturbation Theory : The Spin Orbit Correction [ ] ( ) 2 ℏ2 1 휕 휕 L̂ 2 Zq ⃗ ⃗ ̂ = − r2 − − e + 휉(r)L̂ ⋅ Ŝ 2 휕 휕 2 2 휋휖 2me r r r r ℏ 4 0r ⏟⏞⏟⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ̂ VSO ̂ (0) ⃗ ⃗ ⃗ ̂ (0) ̂ ̂ 2 ̂ 2 ̂ ̂ ̂ Can show that and VSO commute with L , S , and J = L + S Since ⃗ ⃗ 1 ( ) J2 = (L⃗ + S⃗) ⋅ (L⃗ + S⃗) = L2 + S2 + 2L⃗ ⋅ S⃗ rearranges to L̂ ⋅ Ŝ = Ĵ2 − L̂ 2 − Ŝ 2 2 ̂ we can rewrite VSO as 1 [ ] V̂ = 휉(r) Ĵ2 − L̂ 2 − Ŝ 2 SO 2 in terms of vector lengths, Ĵ2, L̂ 2, and Ŝ 2, which are conserved.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction ̂ Calculate 1st-order energy correction to H-atom due to VSO Starting with ⃗ ⃗ 1 [ ] ̂ = ̂ (0) + 휉(r)L̂ ⋅ Ŝ = ̂ (0) + 휉(r) Ĵ2 − L̂ 2 − Ŝ 2 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟2 ̂ (1) ̂ =VSO 1st-order energy correction is [ ][ ] [ ] (1) 휓 ∗ ̂ 휓 휏 1 ∗ 휉 2 ∗ 휒 ∗ ̂ 2 ̂ 2 ̂ 2 휒 휃 휃 휙 E = ,퓁, , V ,퓁, , d = R ,퓁(r) (r)R ,퓁(r)r dr Y퓁, J − L − S Y퓁, sin d d SO ∫ n m퓁 ms SO n m퓁 ms ∫ n n ∫ ∫ m퓁 ms m퓁 ms V 2 1 [ ] = ⟨휉(r)⟩ j(j + 1)ℏ2 − 퓁(퓁 + 1)ℏ2 − s(s + 1)ℏ2 2

휁 ℏ2 ⟨휉 ⟩ Using n,퓁 = (r) we obtain hc0 [ ] (1) 1 E = hc 휁 ,퓁 j(j + 1) − 퓁(퓁 + 1) − s(s + 1) n,j,퓁,s 2 0 n as 1st-order spin-orbit energy correction for H-atom. P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction

[ ] (1) 1 E = hc 휁 ,퓁 j(j + 1) − 퓁(퓁 + 1) − s(s + 1) n,j,퓁,s 2 0 n

[ ( ) ( )] For 퓁 = 0 (s orbitals) and s = 1 → j = 1 and E(1) = 1 1 + 1 − 0 − 1 1 + 1 = 0. 2 2 1 1 2 2 2 2 n, ,0, 2 2

s orbitals are unaffected by the spin-orbit coupling.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction [ ] (1) 1 E = hc 휁 ,퓁 j(j + 1) − 퓁(퓁 + 1) − s(s + 1) n,j,퓁,s 2 0 n For 퓁 = 1 (p orbital) there will be j = 1 − 1 = 1 and j = 1 + 1 = 3 . So [ ( 2 ) 2 ( 2)] 2 (1) 1 3 3 1 1 1 3 E = hc 휁 , + 1 − 2 − + 1 = hc 휁 , , for j = 3 1 0 n 1 2 2 2 2 0 n 1 2 n, ,1, 2 2 2 2 [ ( ) ( )] (1) 1 1 1 1 1 1 E = hc 휁 , + 1 − 2 − + 1 = −hc 휁 , , for j = 1 1 0 n 1 2 2 2 2 0 n 1 2 n, ,1, 2 2 2

Coupling between orbital and spin angular momentum makes specification of electron state with just n, 퓁 , m퓁, and ms insufficient. Also need to specify j. P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom : Spin Orbit Interaction

Write 1st-order spin-orbit coupling energy correction in terms of unperturbed H-atom energy

Combining [ ] (1) 1 E = hc 휁 ,퓁 j(j + 1) − 퓁(퓁 + 1) − s(s + 1) , n,j,퓁,s 2 0 n

훼2 4 Z2휇q4 R∞Z (0) e 휁 ,퓁 = , and E = n 3퓁 퓁 퓁 n 휋2휖 ℏ2 2 n ( + 1)( + 1∕2) 32 0 n

we obtain ( ) 2 (0) { } En n[j(j + 1) − 퓁(퓁 + 1) − s(s + 1) E(1) = SO m c2 퓁(퓁 + 1 )(퓁 + 1) e 0 2

where En is unperturbed energy of H-atom.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Fine Structure of the H-Atom: All Together Executive Summary

Adding in spin orbit and relativistic correction the total energy becomes

(0) [ ( )] E 2 (0) (1) (1) 1 훼 n 3 E , = E + E + E = 1 + − n j n SO r n2 n2 j + 1∕2 4

where j is the total angular momentum (orbit + spin) quantum number.

Our treatment of fine structure stops here.

One could go further with more energy refinements coming from couplings to nuclear degrees of freedom, and even couplings to the quantized electromagnetic field.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Term Symbols

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Term Symbols Atomic spectroscopists have devised approach for specifying set of states having particular set of 퓁, s, and j with a term symbol. Given 퓁, s, and j for electron term symbol is defined as 2s+1 퓁 { letter}j

2s + 1 is called spin multiplicity 퓁 letter is assigned based on numerical value of 퓁 according to

퓁 = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ← numerical value {퓁 letter} ≡ SPDFGHIKLMOQRT ← letter continuing afterwards in alphabetical order. Example What is term symbol for 2 states with j = 1 for p-electron? 2 Solution: 퓁 = 1 has symbol P, and spin multiplicity is 2s + 1 = 2 ⋅ 1 + 1 = 2. 2 2 Thus we have P1∕2 (pronounced “doublet P half.”) P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Term Symbols and Orbital Diagrams

− 퓁 In orbital diagram, e is placed in box associated with and m퓁 with arrows indicating ms value.

Possible orbital diagrams for 1s1 which has 퓁 = 0, s = 1∕2, and j = 1∕2 are

Remember that we can add only one vector component of angular momentum vector in QM—conventionally it is the z component, so mj = m퓁 + ms.

2 Both these orbital diagrams are associated with term symbol S1∕2 (pronounced “doublet S half.”)

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Example How are orbital diagrams for 2p1 configuration associated with term symbols?

Solution: Case of 2p1 with 퓁 = 1 and s = 1∕2 can have j = 1∕2, 3∕2. 2 2 This leads to possible terms of P1∕2 and P3∕2,

We can only associate 2 orbital diagrams with states j = 3∕2, mj = ±3∕2. Other states are part of linear 2 combinations belonging to P1∕2 2 and P3∕2. More systematic approach uses Clebsch-Gordon series.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin H-like Atom Transition Selection Rules

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin H-like Atom Transition Selection Rules For transitions between H-like atom energy states through absorption and emission of light, we require non-zero electric dipole transition moment,

∗ ∗ ⃗ ⟨ ⃗휇⟩ ,퓁, , , ′,퓁′, ′ , ′ = 휓 (r, 휙, 휃)휒 ̂휇휓 ′,퓁′, ′ (r, 휙, 휃)휒 ′ d휏 n m퓁 ms n m퓁 m ∫ n,퓁,m퓁 m n m퓁 m s V s s For H-like atom, is ∑ ⃗̂휇 ⃗̂ ⃗̂ ⃗̂ = qir = Zqerp − qere i ⃗̂ ⃗̂ ⃗̂ Since re = r + rp we can write ⃗̂휇 ⃗̂ ⃗̂ ⃗̂ ⃗̂ ⃗̂ = Zqerp − qe(r + rp) = (Z − 1)qerp − qer Substituting this into transition moment integral gives

∗ ∗ ⃗ ⟨ ⃗휇⟩ ,퓁, , , ′,퓁′, ′ , ′ = 휓 (r, 휙, 휃)휒 (Z − 1)q r̂ 휓 ′,퓁′, ′ (r, 휙, 휃)휒 ′ d휏 n m퓁 ms n m퓁 m ∫ n,퓁,m퓁 m e p n m퓁 m s V s s ∗ ∗ ⃗ − 휓 (r, 휙, 휃)휒 q r̂ 휓 ′,퓁′, ′ (r, 휙, 휃)휒 ′ d휏 ∫ n,퓁,m퓁 m e n m퓁 m V s s

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin H-like Atom Transition Selection Rules :  0    ⟨ ⃗휇⟩ 휓∗ , 휙, 휃 휒∗ ⃗̂ 휓 , 휙, 휃 휒 휏 n,퓁,m ,m ,n′,퓁′,m′ ,m′ = ,퓁, (r ) (Z − 1)qerp n′,퓁′,m′ (r ) m′ d 퓁 s 퓁 s ∫ n m퓁 ms 퓁 s V    ∗ ∗ ⃗ − 휓 (r, 휙, 휃)휒 q r̂ 휓 ′,퓁′, ′ (r, 휙, 휃)휒 ′ d휏 ∫ n,퓁,m퓁 m e n m퓁 m V s s

⃗̂ 1st integral involving rp goes to zero for all transitions 2nd integral gives transition dipole moment

∗ ∗ ⃗ ⟨ ⃗휇⟩ ,퓁, , , ′,퓁′, ′ , ′ = −q 휓 (r, 휙, 휃)휒 r̂ 휓 ′,퓁′, ′ (r, 휙, 휃)휒 ′ d휏 n m퓁 ms n m퓁 m e ∫ n,퓁,m퓁 m n m퓁 m s V s s

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin H-like Atom Transition Selection Rules Transition dipole moment integral is

∗ ∗ ⃗ ⟨ ⃗휇⟩ ,퓁, , , ′,퓁′, ′ , ′ = −q 휓 (r, 휙, 휃)휒 r̂ 휓 ′,퓁′, ′ (r, 휙, 휃)휒 ′ d휏 n m퓁 ms n m퓁 m e ∫ n,퓁,m퓁 m n m퓁 m s V s s

Express ⃗r in spherical coordinates ( ) ⃗ 휃 휙 ⃗ 휃 휙 ⃗ 휃 ⃗ r = r sin cos ex + sin sin ey + cos ez Selection rules for hydrogen-like atom are

Δn = any positive or negative integer Δ퓁 = ±1, ̂휇 , Δm퓁 = 0 for z ̂휇 ̂휇 , Δm퓁 = ±1 for x and y . Δms = 0

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Energy level diagram and transitions of hydrogen atom

Allowed transitions shown as green lines. Energy levels labeled by “terms”, taking both relativistic and spin-orbit energy corrections into account.

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Web Apps by Paul Falstad

Atomic dipole transitions

P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin