Chapter 19 - Magnetism, Angular Momentum, and …P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Biot-Savart Law Jean-Baptiste Biot and Félix Savart worked out

Chapter 19Magnetism, Angular Momentum, and Spin

P. J. Grandinetti

Chem. 4300

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

In 1820 Hans Christian Ørsted discovered that electric current produces a magnetic field thatdeflects compass needle from magnetic north, establishing first direct connection betweenfields of electricity and magnetism.


Biot-Savart LawJean-Baptiste Biot and Félix Savart worked out that magnetic field, B, produced at distance raway from section of wire of length dl carrying steady current is

dB =𝜇04𝜋

dl × rr3 Biot-Savart law

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

current


Microscopic Origins of MagnetismShortly after Biot and Savart, Ampére suggested that magnetism in matter arises from amultitude of ring currents circulating at atomic and molecular scale.

André-Marie Ampére1775 - 1836


Magnetic dipole moment from current loopCurrent flowing in flat loop of wire with area A will generate magnetic fieldwhich, at distance much larger than radius, r, appears identical to fieldproduced by point magnetic dipole with strength of

𝜇 = |𝜇| = ⋅ Aradius

current

magneticdipole

ExampleWhat is magnetic dipole moment induced by e− in circular orbit of radius r with linear velocity v?

Solution: For e− with linear velocity of v the time for one orbit is torbit = 2𝜋r∕v.Taking current as charge, −qe, passing point per unit time we have

= −qe∕torbit = −qev2𝜋r

Taking area of current loop as A = 𝜋r2 we obtain magnetic dipole moment of

𝜇 = A =(−qev

2𝜋r

)(𝜋r2) = −

qe2

r v


Electron Orbital Magnetic DipoleRecalling 𝜇 = −

qe2

r v, for electron with angular momentum, rearrange L = mr × v, to get

Lm

= r × v and then obtain 𝜇 = −qe2

Lme

= −𝛾LL where 𝛾L ≡ qe∕(2me)

𝛾L is called orbital gyromagnetic ratio : ratio of magnetic moment to orbital angular momentum

magneticdipole

Angularmomentum

radius

current

Angular momentum vector is antiparallel to magnetic dipolevector due to negative electron charge.

Don’t forget : current flow is defined in opposite direction ofelectron flow.

Despite simple derivation, 𝜇 = −𝛾LL is general and holds fornon-circular orbits as long as angular momentum is conserved.

𝜇 = −𝛾LL is independent of size, period, and even shape oforbit.


Electron Orbital Magnetic Dipole OperatorFrom e− orbital motion the magnetic dipole moment operator is

𝜇𝓁 = −qe

2me

L = −𝜇BℏL

𝜇B is defined as Bohr magneton (an atomic unit for dipole moments) and given by

𝜇B ≡ qeℏ2me

= 9.274009992054043 × 10−24 J/T

Magnitude of dipole moment is 𝜇𝓁 = 𝜇B√𝓁(𝓁 + 1)

Operator for z component of electron orbital magnetic dipole moment is

��z = −𝜇B

ℏLz

For z (or x or y) components only observed with quantized values of

𝜇z = −𝜇Bℏ⟨Lz⟩ = −

𝜇Bℏ(m𝓁ℏ) = −𝜇Bm𝓁


Classical magnetic dipole in a uniform magnetic fieldEnergy of classical magnetic dipole in uniform magnetic field is

Potential energy is

V = −𝜇 ⋅ B = −|𝜇||B| cos 𝜃 N

Sz

x

Note: 𝜃 = 180◦ is an unstable equilibrium point, i.e., potential is a maximum at 𝜃 = 180◦.Take kinetic energy as K = 1

2 I𝜔2, then total energy is

E = 12

I𝜔2 − 𝜇 ⋅ B = 12

I𝜔2 − |𝜇||B| cos 𝜃With the magnetic dipole and magnetic field vectors in the x-z plane, magnetic dipole experiencestorque about y given by

𝜏 = 𝜇 × B =[|𝜇||B| sin 𝜃] ey

In absence of friction total energy remains constant and compass needle oscillates about direction ofmagnetic field.


Magnetic dipole with angular momentum

Spinning top precesses in gravitational field.

Precession Direction

dJdt

= 𝜏 = R × Mg

Magnetic dipole with angular momentumprecesses in magnetic field

N

S

dJdt

= 𝜏 = 𝜇 × B


e−’s magnetic dipole from orbital motion in uniform magnetic fieldZeeman interaction

For e− with orbital motion substitute 𝜇𝓁 = −𝜇BℏL into V = −𝜇 ⋅ B and obtain

V = −𝜇BℏL ⋅ B

Coupling of e−’s magnetic dipole moment to external magnetic field is called Zeemaninteraction.

Taking magnetic field as along z axis, B = Bzez, expression simplifies to

V = −𝜇Bℏ

LzBz

Zeeman interaction lifts degeneracy of m𝓁 levels of electron in H-atom.


Classical magnetic dipole with angular momentum in uniform magnetic fieldPlace classical magnetic dipole with anti-parallel angular momentum in magnetic field and magneticmoment experiences torque

𝜏 = 𝜇 × B = −𝛾LL × B = 𝛾LB × L

Torque causes change in angular momentum. Newton’s 2nd law says

dLdt

= 𝜏 = 𝛾LB × L

Taking B = Bzez, and definingΩ0 = −𝛾LBz

we write cross product in determinant form

dLdt

=|||||||

ex ey ez0 0 −Ω0Lx Ly Lz

||||||| = LyΩ0ex − LxΩ0ey

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

dLxdt

= Ω0Ly,dLy

dt= −Ω0Lx,

dLz

dt= 0

Coupled differential eqs. involving Lx and Ly can be decoupled by defining

L± = Lx ± iLy

Then we finddL+

dt=

dLxdt

+ idLy

dt= Ω0Ly − iΩ0Lx = −iΩ0(Lx + iLy) = −iΩ0L+

dL−dt

=dLxdt

− idLy

dt= Ω0Ly + iΩ0Lx = iΩ0(Lx − iLy) = iΩ0L−

ordL±

dt= ∓iΩ0L± whose solution is L±(t) = L±(0)e∓iΩ0t

Converting L± back to Cartesian and taking 𝜇 = −𝛾LL we find ...P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin

Classical magnetic dipole with angular momentum in uniform magnetic field

𝜇x(t) = 𝜇x(0) cosΩ0t + 𝜇y(0) sinΩ0t𝜇y(t) = 𝜇y(0) cosΩ0t − 𝜇x(0) sinΩ0t𝜇z(t) = 𝜇z(0)

These equations for magnetic dipole vector componentsdescribe precession motion of vector around magneticfield direction at angular frequency of Ω0.When magnetic dipole vector has angular momentumtorque does not cause dipole to align with B, but rathercauses it to precess about direction of B.


Classical Magnetic dipole in a Non-Uniform Magnetic Field



If magnetic dipole is in non-uniformmagnetic field, B(r), then forces acting on2 poles will not cancel and dipole starts totranslate along magnetic field gradientdirection.In this case net force is

Fnet = F+ + F− = ∇(𝜇 ⋅ B(r)

)Dipole oriented in magnetic field gradientis pulled up.

N

S



Differently oriented dipole in same magneticfield gradient is pulled down.

N

S



When dipole is perpendicular to lines of forcethere is no translation of dipole. N S



N

S

N S

N

S

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


Magnetic dipoles with angular momentumOn the other hand, if the magnetic dipoles also have angular momentum...

N

S

N S

N

S

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


Classical Magnetic dipole in a Non-Uniform Magnetic FieldNow with Angular MomentumTake example of magnetic field given by

B(r) = −gxex + (B0 + gz)ez

g is gradient strength. Maxwell’s equations require ∇ ⋅ B = 0, so need gradient along x tosatisfy this constraint. Thus we find

Fnet = −∇V = ∇(𝜇 ⋅ B(r)

)= ∇

(𝜇 ⋅ (−gxex + (B0 + gz)ez

)= ∇

(−gx𝜇xex + (B0 + gz)𝜇zez

)= −g𝜇xex + g𝜇zez

Unlike compass needle, we want to consider magnetic dipole with angular momentum. Thus,𝜇 precesses about magnetic field direction while translating. For this precession motion we use

𝜇x(t) = 𝜇x(0) cosΩ0t + 𝜇y(0) sinΩ0t𝜇y(t) = 𝜇y(0) cosΩ0t − 𝜇x(0) sinΩ0t𝜇z(t) = 𝜇z(0)


Classical Magnetic dipole in a Non-Uniform Magnetic FieldNow with Angular MomentumUsing

Fnet = −g𝜇x(t)ex + g𝜇z(t)ez

assume B0 ≫ gx, gz and use 𝜇(t) for precessing magnetic dipole moment to obtain

Fnet(t) = −g(𝜇x(0) cosΩ0t + 𝜇y(0) sinΩ0t

)⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

Fx(t)

ex + g𝜇z(0)⏟⏟⏟

Fz

ez

Time independent Fz component of force will push dipole in z direction—which way depends oninitial orientation of dipole.Time dependent Fx(t) component of force will push dipole back and forth about its originalposition with a time averaged displacement of zero along x.Bottom line: passing magnetic dipole through magnetic field gradient along zprovides means of measuring 𝜇z(0).Basis for seminal experiment in 1922 by Otto Stern and Walther Gerlach.


The Stern-Gerlach Experiment - First Some BackgroundIn 1896 Pieter Zeeman discovered clue about magnetic properties of yet unexplainedconstituents of atom when he reported broadening of yellow D-lines of sodium in flamewhen held inside strong magnetic field. Became known as Zeeman effect.

Bohr’s 1913 theory of atom predicted that e− orbital angular momentum would bequantized.In 1916 Arnold Sommerfeld refined Bohr’s theory to allow for elliptical orbits described by3 quantum numbers: n, k, and m, where

▶ n = 1, 2, 3,…, called the principal quantum number,▶ k = 1,… , n, called the azimuthal quantum number▶ m = −k,−k + 1,… ,m ≠ 0,… , k − 1, k, called the magnetic quantum number.

m described quantization of z-component of e−’s orbital angular momentum.

Refinements allowed Sommerfeld to explain Zeeman effect through dependence of orbitalmagnetic dipole moment on magnetic quantum number.


The Stern-Gerlach Experiment

In 1922 Otto Stern and Walther Gerlach designed apparatus in attempt to measure thisquantization with beam of neutral Ag atoms.

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

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

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


Schematic diagram of the Stern Gerlach apparatus.

S

N

Measurementresult

Classicalprediction

OvenCollimator

Atom beam

Classical prediction is all values of 𝜇z between −|𝜇| and |𝜇| 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.


“Attached is the experimental proof of directionalquantization. We congratulate you on theconfirmation of your theory.”

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


The Stern-Gerlach ExperimentStern-Gerlach experiment in 1922 was before De Broglie’s hypothesis in 1923 and Schrödinger’swave 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 appreciatefull significance of their discovery.In 1927 Phipps & Taylor used S-G technique on beam of H atoms. In 1927 Schrödinger’s waveequation and solutions for H atom was known.Orbital angular momentum of H atom ground state (n = 1, 𝓁 = 0, and m𝓁 = 0) predicts only 1outcome for z component of e−’s magnetic dipole moment, 𝜇z = 0.

N

S

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


Electron Spin

Hypothesis (now theory) that explains Stern-Gerlach (and Phipps & Taylor) result is that e− hasintrinsic (built-in) magnetic dipole moment, 𝜇s, arising from intrinsic angular momentum, S, calledspin.

Property called spin because if e− is charged ball spinning about its own axis then it would havemagnetic 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 momentumquantum number.

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


Electron Spin and the Dirac Wave Equation

In 1929 Paul Dirac (1902-1984) proposed relativistic wave equation for e−, showing it must haveintrinsic angular momentum of s=1/2 and magnetic moment of

𝜇s = −gs𝜇Bℏ

S where gs = 2.00231930436182 is called electron g-factor

Magnitude of spin dipole moment is 𝜇s = gs𝜇B√

s(s + 1)In same theory he also predicted existence of anti-electron (and anti-matter).


Spin OrbitalsInstead of switching to Dirac equation we modify (ad hoc) Schrödinger equation solutions tobe product wave function with new degree of freedom

𝜓(r)𝜒(S)

𝜒(S) is spin part of e−’s wave function whose stationary states, 𝜒ms, are given symbols 𝛼 for

ms = +1∕2 and 𝛽 for ms = −1∕2.Spin part is expressed as linear combination

𝜒(S) = c𝛼𝛼 + c𝛽𝛽

c𝛼 and c𝛽 are complex coefficients.Two spin states, 𝛼 and 𝛽, are orthogonal,

∫ 𝛼∗𝛼d𝜎 = ∫ 𝛽∗𝛽d𝜎 = 1, ∫ 𝛼∗𝛽d𝜎 = ∫ 𝛽∗𝛼d𝜎 = 0


Spin Orbitals

H-atom stationary states modified toinclude e− spin state

𝜓n,𝓁,m𝓁 ,ms(r, 𝜃, 𝜙, 𝜎) = Rn,𝓁(r)Y𝓁,m𝓁

(𝜃, 𝜙)𝜒ms

and are called spin orbitals.

H-atom stationary states are specified by4 quantum numbers n, 𝓁, m𝓁, and ms.

Write H-atom stationary states withshorthand notation.

𝜓n,0,0, 12≡ n s 𝛼, 𝜓n,0,0,− 1

2≡ n s 𝛽,

𝜓n,1,−1, 12≡ n p−1 𝛼, 𝜓n,1,−1,− 1

2≡ n p−1 𝛽,

𝜓n,1,0, 12≡ n p0 𝛼, 𝜓n,1,0,− 1

2≡ n p0 𝛽,

𝜓n,1,1, 12≡ n p1 𝛼, 𝜓n,1,1,− 1

2≡ n p1 𝛽,

𝜓n,2,−2, 12≡ n d−2 𝛼, 𝜓n,2,−2,− 1

2≡ n d−2 𝛽,

𝜓n,2,−1, 12≡ n d−1 𝛼, 𝜓n,2,−1,− 1

2≡ n d−1 𝛽,

𝜓n,2,0, 12≡ n d0 𝛼, 𝜓n,2,0,− 1

2≡ n d0 𝛽,

𝜓n,2,1, 12≡ n d1 𝛼, 𝜓n,2,1,− 1

2≡ n d1 𝛽,

𝜓n,2,2, 12≡ n d2 𝛼, 𝜓n,2,2,− 1

2≡ n d2 𝛽.


Spin Orbitals

Occupancy of H-atom stationary states is often illustrated in form of an orbital diagram.Below is case of n = 1, 𝓁 = 0, m𝓁 = 0, and ms =

12 .

2s1s 2p 3s 3p 3d


Electron Spin and Total Magnetic Dipole Moment OperatorWe complete treatment of e−’s magnetic dipole moment in stating that total magnetic dipolemoment for e− includes both orbital and spin contributions,

𝜇 = 𝜇𝓁 + 𝜇s = −g𝓁𝜇B

ℏL −

gs𝜇Bℏ

S

g𝓁 = 1 and is called orbital g factor.gs = 2.00231930436182 is called electron g-factor.Total magnetic dipole moment operator for e− is

𝜇 = −𝜇B

ℏ

(g𝓁L + gs

S)

and for total z component of the magnetic dipole moment operator we have

��z = −𝜇B

ℏ(g𝓁Lz + gsSz

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

Addition of Angular Momentum


Addition of Angular MomentumDiscovery of electron spin angular momentum means that total electron angularmomentum of H-atom becomes

J = L + S

In classical mechanics this would be calculated trivially,

J =⎛⎜⎜⎝

JxJyJz

⎞⎟⎟⎠ =⎛⎜⎜⎝

LxLyLz

⎞⎟⎟⎠ +⎛⎜⎜⎝

SxSySz

⎞⎟⎟⎠But this requires that we know all 3 vector components of L and S simultaneously.In QM we can only know length of each angular momentum vector and one componentsimultaneously. That is, can only know sum of one component,

Jz = Lz + Sz

which tells us that mj = m𝓁 + ms


Addition of Angular MomentumWhile QM lacks information needed to calculate length of total angular momentumvector, J, we still have constraints on how big or small the total can be.Largest vector sum is when both vectors point in same directionSmallest vector sum is when both point in opposite directions

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


Addition of Angular Momentum|L + S| ≥ |J| ≥ ||||L| − |S|||| translates to√

j(j + 1) ≥ |||√𝓁(𝓁 + 1) −√

s(s + 1)|||constraining j to fall between limits |𝓁 − s| ≤ j ≤ 𝓁 + s with integral or half-integral values.

ExampleWhat are the total angular momenta resulting from the addition of 𝓁 = 0 and s = 1

2?

Solution: j = |||0 − 12||| to |||0 + 1

2||| gives j = 1

2

ExampleWhat are the total angular momenta resulting from the addition of 𝓁 = 1 and s = 1

2?

Solution: j = |||1 − 12||| to |||1 + 1

2||| gives j = 1

2 and 32 .

For each j value there will be 2j + 1 values of mj, that is, mj = −j,−j + 1,… , j − 1, j


Electron Magnetic Moment from Total Angular MomentumTotal magnetic dipole moment operator for e− is

𝜇 = −gj𝜇B

ℏJ

and for total z component of the magnetic dipole moment operator we have

��z = −gj𝜇B

ℏJz

gj is called Landé g factor:

gj =j(j + 1)(g𝓁 + gs) + [𝓁(𝓁 + 1) − s(s + 1)](g𝓁 − gs)

2j(j + 1)Magnetic moment associated with electron having given value of j is

𝜇j = gj𝜇B√

j(j + 1)


Landé g factors for various 𝓁 values with s = 1∕2.

gj =j(j + 1)(g𝓁 + gs) + [𝓁(𝓁 + 1) − s(s + 1)](g𝓁 − gs)

2j(j + 1)

𝓁 s j gj0 1/2 1/2 2.0023193043618211 1/2 1/2 0.6658935652127271 1/2 3/2 1.3341064347872732 1/2 3/2 0.7995361391276362 1/2 5/2 1.2004638608723643 1/2 5/2 0.8568115279483113 1/2 7/2 1.1431884720516884 1/2 7/2 0.8886311884042424 1/2 9/2 1.111368811595758


Clebsch-Gordon series expansion

How do the wave functions associated with 𝓁, m𝓁, s, and ms combine to form the wavefunctions associated with j and mj?

They are related through Clebsch-Gordon series expansion,

𝜓j,mj=

𝓁∑m𝓁=−𝓁

s∑ms=−s

C(j,mj,𝓁,m𝓁, s,ms)𝜓𝓁,m𝓁 ,s,ms

C(j,mj,𝓁,m𝓁, s,ms) are called Clebsch-Gordon coefficients.

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

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


Clebsch-Gordon coefficients

j 1j 2m1m2|j 1j 2JM

= ( − 1) J − j 1− j 2 j 2j 1m2m1|j 2j 1JMd

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

ExampleConstruct wave function with j = 1

2 and mj = − 12 for p-electron in H-atom in terms of wave functions

associated with 𝓁, m𝓁, s, and ms.

Solution: Addition of 𝓁 = 1 and s = 12 leads to 2 possibilities of j = 1

2 and j = 32 . Focus on j = 1

2 statesand write Clebsch-Gordon series as

𝜓n,j= 12 ,mj

=𝓁∑

m𝓁=−𝓁

s∑ms=−s

C(

j = 12 ,mj,𝓁 = 1,m𝓁 , s =

12 ,ms

)𝜓

n,𝓁=1,m𝓁 ,s=12 ,ms

Expanding out the mj = − 12 case we find six terms:

𝜓n, 12 ,−12

= C(

12 , −

12 , 1,−1, 1

2 ,−12

)n p−1𝛽 + C

(12 , −

12 , 1, 0, 1

2 ,−12

)n p0𝛽

+ C(

12 , −

12 , 1, 1, 1

2 ,−12

)n p1𝛽 + C

(12 , −

12 , 1,−1, 1

2 ,12

)n p−1𝛼

+ C(

12 , −

12 , 1, 0, 1

2 ,12

)n p0𝛼 + C

(12 , −

12 , 1, 1, 1

2 ,12

)n p1𝛼

Only CC coefficients with mj = m𝓁 + ms = − 12 are non-zero.


ExampleConstruct wave function with j = 1

2 and mj = − 12 for p-electron in H-atom in terms of wave functions

associated with 𝓁, m𝓁, s, and ms.

Solution: We are left with

𝜓n, 12 ,−12= C

(12 , −

12 , 1, 0, 1

2 ,−12

)n p0𝛽 + C

(12 , −

12 , 1,−1, 1

2 ,12

)n p−1𝛼

Use 1 × 12 Clebsch-Gordon Coefficient table to find

𝜓n, 12 ,−12= 1√

3n p0𝛽 −

√23

n p−1𝛼

Using orbital diagrams we could represent j = 12 , mj = − 1

2 state as


Addition of three or more angular momenta

What about addition of three or more angular momenta?

For example, What about 𝓁 = 1, s = 12 , and i = 1

2?

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

First add 𝓁 and s to get j = |𝓁 − s| to 𝓁 + 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 𝓁 to get f = |j − 𝓁| to j + 𝓁.

or

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


ExampleWhat are total angular momenta values and number of states that result when 𝓁 = 1, s = 1

2 ,and i = 1

2 are added?

Solution: Starting with 𝓁 and s we find thatFirst, for 𝓁 = 1 and s = 1

2 we find j = |𝓁 − s| to 𝓁 + s gives j = 12 , 3

2

then, for j = 12 and i = 1

2 we find that f = |j − i| to j + i gives f = 0, 1

and for j = 32 and i = 1

2 we find that f = |j − i| to j + i gives f = 1, 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


Fine Structure of the H-Atom

1 Relativistic Correction:(1) = −

p4

8m3c30

2 Spin Orbit Interaction:VSO = 𝜉(r)L ⋅ S

where 𝜉(r) is the spin-orbit coupling constant.


Fine Structure of the H-Atom : Relativistic CorrectionWhile classical kinetic energy expression is good starting approximation for H-atom there areslight relativistic effects that can be observed in emission spectra of H-atoms.

Instead of K =p2

2m, use K =

mc20√

1 − (v∕c0)2− mc2

0

Derivation in notes shows relativistic correction of e− kinetic energy gives

(1) = −p4

8m3c30

This, in turns, gives a 1st-order correction to energy

E(1)r =

E2n

mec20

{4n

𝓁 + 1∕2− 3

}Note: relativistic correction makes energy dependent on 𝓁.


Fine Structure of the H-Atom : Spin Orbit InteractionWhen e− has orbital angular momentum the e−’s intrinsic spin magnetic dipole, 𝜇s interactswith 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 andcalculate spin-orbit interaction as

VSO = −𝜇s ⋅ Borb

Lengthy derivation in notes gives ...P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin

Fine Structure of the H-Atom : Spin Orbit InteractionLengthy derivation in notes gives

VSO = 𝜉(r)L ⋅ S

𝜉(r) is spin orbit coupling constant, given by

𝜉(r) = 12m2

ec20

1r

d𝜙(r)dr

𝜙(r) is electrostatic potential at electron.

Convention is to report spin orbit coupling constant as

𝜁n,𝓁 = ℏ2

hc0 ∫ 𝜓∗n,𝓁,m𝓁

𝜉(r)𝜓n,𝓁,m𝓁d𝜏 = ℏ2

hc0⟨𝜉(r)⟩

𝜁n,𝓁 is a wavenumber and hc0𝜁n,𝓁 is an energy.


Fine Structure of the H-Atom : Spin Orbit InteractionFor H-like atom we can further calculate

𝜁n,𝓁 =𝛼2R∞Z4

n3𝓁(𝓁 + 1)(𝓁 + 1∕2)

𝛼 is the fine structure constant

𝛼 ≡ q2e

4𝜋𝜖0ℏc0= 0.007297352566206478

R∞ is Rydberg constant,

R∞ =meq4

e

8𝜖20h3c0

Note that RH = R∞∕(1 + me∕mp)

For 2p electron in H-atom we have 𝜁2,1 = 𝛼2R∞∕24 = 0.24 cm−1 or 30.2 𝜇eV, which is a smallshift on E2 = −3.4 eV, the 2p energy.Such small perturbation is observable in emission spectrum of H-atoms.VSO plays a bigger role in heavier elements since 𝜁n,𝓁 ∝ Z4.


Perturbation Theory : The Spin Orbit CorrectionLet’s calculate the 1st-order energy correction from VSO using the H-atom (0) as

(0) = − ℏ2

2me

[1r2𝜕𝜕r

(r2 𝜕𝜕r

)− L2

r2ℏ2 −Zq2

e4𝜋𝜖0r

]

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

Hamiltonian above commutes with L, i.e., [, L] = 0, so orbital angular momentum is conserved.

If spin-orbit coupling didn’t exist then [, S] = 0 and spin angular momentum is conserved.But once we include spin-orbit coupling to = (0) + VSO,

VSO = 𝜉(r)L ⋅ S

then we find [, L] ≠ 0 and [, S] ≠ 0.In other words orbital and spin angular momentum are no longer separately conserved whenspin-orbit coupling is present.


Perturbation Theory : The Spin Orbit Correction

= − ℏ2

2me

[1r2𝜕𝜕r

(r2 𝜕𝜕r

)− L2

r2ℏ2 −Zq2

e4𝜋𝜖0r

]⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

(0)

+ 𝜉(r)L ⋅ S⏟⏞⏟⏞⏟

VSO

Can show that (0) and VSO commute with L2, S2, and J = L + SSince

J2 = (L + S) ⋅ (L + S) = L2 + S2 + 2L ⋅ S

it will be helpful to rewrite VSO using

L ⋅ S = 12(J2 − L2 − S2)

asVSO = 1

2𝜉(r)

[J2 − L2 − S2]

in terms of quantities J2, L2, and S2, which are conserved.P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin

Fine Structure of the H-Atom : Spin Orbit InteractionCalculate 1st-order energy correction to H-atom due to VSOStarting with

= (0) + 𝜉(r)L ⋅ S = (0) + 12𝜉(r)

[J2 − L2 − S2]

1st-order energy correction is

E(1)SO = ∫V

𝜓∗n,𝓁,m𝓁 ,ms

VSO𝜓n,𝓁,m𝓁 ,msd𝜏

= 12

[∫ R∗

n,𝓁(r)𝜉(r)Rn,𝓁(r)r2dr] [

∫ ∫ Y∗𝓁,m𝓁

𝜒∗ms

[J2 − L2 − S2]Y𝓁,m𝓁

𝜒mssin 𝜃d𝜃d𝜙

]= 1

2⟨𝜉(r)⟩ [j(j + 1)ℏ2 − 𝓁(𝓁 + 1)ℏ2 − s(s + 1)ℏ2]

Using 𝜁n,𝓁 = ℏ2

hc0⟨𝜉(r)⟩ we obtain

E(1)n,j,𝓁,s =

12

hc0𝜁n,𝓁[j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)

]as 1st-order spin-orbit energy correction for H-atom.


Fine Structure of the H-Atom : Spin Orbit Interaction

E(1)n,j,𝓁,s =

12

hc0𝜁n,𝓁[j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)

]

For 𝓁 = 0 (s orbitals) and s = 12 → j = 1

2 and E(1)

n,12 ,0,

12

=[

12

(12 + 1

)− 0 − 1

2

(12 + 1

)]= 0.

s orbitals are unaffected by the spin-orbit coupling.


Fine Structure of the H-Atom : Spin Orbit Interaction

E(1)n,j,𝓁,s =

12

hc0𝜁n,𝓁[j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)

]For 𝓁 = 1 (p orbital) there will be j = 1 − 1

2 = 12 and j = 1 + 1

2 = 32 . So

E(1)

n, 32 ,1,12

= 12

hc0𝜁n,1

[32

(32 + 1

)− 2 − 1

2

(12 + 1

)]= 1

2hc0𝜁n,1, for j = 3

2

E(1)

n, 12 ,1,12

= 12

hc0𝜁n,1

[12

(12 + 1

)− 2 − 1

2

(12 + 1

)]= −hc0𝜁n,1, for j = 1

2

Coupling between orbital and spin angular momentum makes specification of electron state with just n,𝓁, m𝓁, and ms insufficient. Also need to specify j.


Fine Structure of the H-Atom : Spin Orbit InteractionWe can also write 1st-order energy correction from spin-orbit coupling in terms of unperturbedenergy of H-atom.

CombiningE(1)

n,j,𝓁,s =12

hc0𝜁n,𝓁[j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)

],

𝜁n,𝓁 =𝛼2R∞Z4

n3𝓁(𝓁 + 1)(𝓁 + 1∕2), and E(0)

n =Z2𝜇q4

e

32𝜋2𝜖0ℏ2n2

we obtain

E(1)SO =

(E(0)

n

)2

mec20

{n[j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)

𝓁(𝓁 + 12 )(𝓁 + 1)

}where En is unperturbed energy of H-atom.


Fine Structure of the H-Atom: All TogetherExecutive Summary

Adding in spin orbit and relativistic correction the total energy becomes

En,j = E(0)n + E(1)

SO + E(1)r =

E(0)1

n2

[1 + 𝛼2

n2

(n

j + 1∕2− 3

4

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

Our treatment of hydrogen atom fine structure stops here.

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


Term Symbols


Term SymbolsAtomic spectroscopists have devised approach for specifying set of states having particular set of 𝓁, s,and j with a term symbol.Given 𝓁, s, and j for electron term symbol is defined as

2s+1 {𝓁 letter}j

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

𝓁 = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 ← numerical value{𝓁 letter} ≡ S P D F G H I K L M O Q R T ← letter

continuing afterwards in alphabetical order.

ExampleWhat is term symbol for 2 states with j = 1

2 for p-electron?

Solution: 𝓁 = 1 has symbol P, and spin multiplicity is 2s + 1 = 2 ⋅ 12 + 1 = 2.

Thus we have 2P1∕2 (pronounced “doublet P half.”)P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin

Term Symbols and Orbital Diagrams

In orbital diagram e− is placed in box associated with 𝓁 and m𝓁 with arrows indicating msvalue.

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

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

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


ExampleHow are orbital diagrams for 2p1 configuration associated with term symbols?

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

We can only associate 2 orbitaldiagrams with states j = 3∕2,mj = ±3∕2.

Other states are part of linearcombinations belonging to 2P1∕2and 2P3∕2.

More systematic approach usesClebsch-Gordon series.


Energy levels of the hydrogen atom.

Taking both relativistic and spin-orbit energycorrections into account we arrive at energylevel diagram of hydrogen labeled by “terms”


Nuclear Spin Angular Momentum


Nuclear SpinIn 1924 Pauli suggested existence of proton spin and magnetic dipoleIn 1933 Otto Stern discovers that proton has intrinsic spin angular momentum and associatedintrinsic magnetic dipole moment.Like e−, proton spin quantum number is I = 1∕2 with mI = ±1∕2.Magnitude of proton magnetic dipole is much smaller than e−’s.

Like e− one might expect proton magnetic dipole to be 𝜇I =qe

2mpI

Key difference is division by proton instead of e− mass.Mass ratio predicts proton dipole moment is mp∕me ≈ 1836 times smaller than e− but it is only656 times smaller than e−’s.Define nuclear magneton as

𝜇N ≡ qeℏ2mp

= 5.050783698211084 × 10−27 J/T,

and proton magnetic dipole moment is 𝜇p = 2.792847351193473𝜇N


Nuclear Spin

Soon after Rutherford proposed his model of atom with e− orbiting positively charged nucleus itwas 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 toprotons, and called them neutrons.

In 1932 James Chadwick produced 1st convincing evidence of neutron’s existence.

In 1937 Isador Rabi and his group at Columbia University developed nuclear magnetic resonance(NMR) method to measure magnetic dipole moments of atomic nuclei.

Rabi found that neutron spin quantum number is I = 1∕2 and its magnetic dipole moment is

𝜇n = −1.913042723136664𝜇N

Opposite in sign and little smaller than proton magnetic dipole, 𝜇p = 2.792847351193473𝜇N


Nuclear Spin

Total angular momentum, I, of nuclear ground state is determined by number of protonsand neutrons in nucleus.

Nuclear magnetic dipole moments are related to I according to

𝜇z = 𝛾II,

𝛾I is called nuclear gyromagnetic ratio.


Nuclear magnetic dipole moments of various (NMR Active) IsotopesZ N Isotope Mass/(g/mol) Abundance/% I Dipole Moment/(𝜇N)1 0 1H 1.00782503207 99.985 1/2 2.792847392 1 3He 3.01602931914 0.000137 1/2 -2.127624853 4 7Li 7.016004548 92.41 3/2 3.25642684 5 9Be 9.012182201 100 3/2 -1.17785 6 11B 11.009305406 80.2 3/2 2.68864896 7 13C 13.00335483778 1.11 1/2 0.70241187 7 14N 14.00307400478 99.634 1 0.4037618 9 17O 16.999131703 0.038 5/2 -1.893799 10 19F 18.998403224 100 1/2 2.62886810 11 21Ne 20.993846684 0.27 3/2 -0.66179711 12 23Na 22.98976928087 100 3/2 2.2175212 13 25Mg 24.985836917 10 5/2 -0.8554513 14 27Al 26.981538627 100 5/2 3.641506914 15 29Si 28.9764947 4.683 1/2 -0.5552915 16 31P 30.973761629 100 1/2 1.131616 17 33S 32.971458759 0.75 3/2 0.643821217 18 35Cl 34.968852682 75.77 3/2 0.821874319 20 39K 38.963706679 93.2581 3/2 0.391466220 23 43Ca 42.958766628 0.135 7/2 -1.31764321 24 45Sc 44.955911909 100 7/2 4.756486622 25 47Ti 46.951763088 7.44 5/2 -0.7884823 28 51V 50.943959507 99.75 7/2 5.1487057324 29 53Cr 52.940649386 9.501 3/2 -0.4745425 30 55Mn 54.938045141 100 5/2 3.453226 31 57Fe 56.935393969 2.119 1/2 0.0904427 32 59Co 58.933195048 100 7/2 4.627


Documents

Chapter 19 - Magnetism, Angular Momentum, and …P. J. Grandinetti Chapter 19: Magnetism, Angular Momentum, and Spin Biot-Savart Law Jean-Baptiste Biot and Félix Savart worked out