Physical Chemistry (II)

Lecture 11

CHEM 3172-80

Lecturer: Hanning Chen, Ph.D.02/27/2017

Spin-Orbit Coupling

Quiz 10

5 minutes

Please stop writing when the timer stops !

Electron Spin

electron

spin quantum number

S = 12

ms = + 12

ms = − 12

α spin

β spin

definition of multiplicity: 2S +1For a one-electron system: what is its multiplicity ?

12× 2 +1= 2 doublet state

For a system with two paired electrons at the same orbital: what is its multiplicity ?

ϕ(r)

(number of available spin states)

Singlet State

Ψ = 1N!

ϕ1 r1( ) ϕ1 r2( ) ... ϕ1 rn( )ϕ2 r1( ) ϕ2 r2( ) ... ϕ2 rn( )... ... ... ...

ϕn r1( ) ϕn r2( ) ... ϕn rn( )

Row : orbital indexColumn: particle index

Ψ = 12

ϕ r1( )α r1( )ϕ r2( )β r2( )−ϕ r2( )α r2( )ϕ r1( )β r1( )⎡⎣ ⎤⎦

= 12ϕ r1( )ϕ r2( ) α r1( )β r2( )− β r1( )α r2( )⎡⎣ ⎤⎦

a subtractive mixture of two spin configurations !

magnitude of the total spin ???

S2 α r1( )β r2( )− β r1( )α r2( )( ) = 0 α r1( )β r2( )− β r1( )α r2( )( )S2 : spin magnitude operator S = 0total spin quantum number:

Sz : spin projection operator

Sz α r1( )β r2( )− β r1( )α r2( )( ) = 0 α r1( )β r2( )− β r1( )α r2( )( )total spin magnetic quantum number: mS = 0

−

Triplet State

S2 α r1( )α r2( )( ) = 2!2 α r1( )α r2( )( ) = S × (S +1)!2 α r1( )α r2( )( )

S2 : spin magnitude operator S = 1total spin quantum number:

Sz : spin projection operator

Sz α r1( )α r2( )( ) = +! α r1( )α r2( )( )total spin magnetic quantum number: mS = +1

ϕ1 ϕ2Ψ = 1

2ϕ1 r1( )α r1( )ϕ2 r2( )α r2( )−ϕ1 r2( )α r2( )ϕ2 r1( )α r1( )⎡⎣ ⎤⎦

= 12

ϕ1 r1( )ϕ2 r2( )−ϕ2 r1( )ϕ1 r2( )⎡⎣ ⎤⎦α r1( )α r2( )

ϕ1 ϕ2 S2 = 2!2 S = 1 mS = −1

+ S2 = 2!2 S = 1 mS = 0 α r1( )β r2( ) + β r1( )α r2( )

addictive superposition

2S +1= 3

Spin-Orbit Coupling

nucleus

!p = m!v

!r

!L = !r × !p

!L : orbital angular momentum

!S

!S: spin angular momentum

total angular momentum !J =!L +!S

maximum magnitude of J: J max = L + S

!L

!S

minimum magnitude of J: J min = L − S

!L

!S

!L

!S

Total Angular Momentum Quantum Number

For a given set of L and S,

J = L − S , L − S +1,...,L + S −1,L + S{ }For a single electron at an s orbital: s1

L = 0 S = 12

J = 12

⎧⎨⎩

⎫⎬⎭

For a single electron at an f orbital: f 1

L = 3 S = 12

J = 52, 72

⎧⎨⎩

⎫⎬⎭

Mangitude of total angular momentum:

J = J J +1( )!

Energy of a Quantum RotorEnergy levels of a three-dimensional rotor:

E ∝ J 2 !J =!L +!S

J2 = J!"

i J!"= ("L +"S) i (

"L +"S) =

"L i"L +"S i"S + 2

"L i"S

!L i!L : orbital rotation

!S i!S : spin rotation

!L i!S : spin-orbit coupling

!L i!L = l(l +1)"2

!S i!S = s(s +1)"2

!J i!J = j( j +1)"2

Orbital-spin coupling energy:

Espin−orbit =

12hc !Α j( j +1)− l(l +1)− s(s +1)( )

!Α :spin-orbit coupling constant

Fine StructureEnergy level splitting:

4 fFor a single electron at 4f orbital: L = 3 S = 12

J = 52, 72

⎧⎨⎩

⎫⎬⎭

Espin−orbit =12hc !Α j( j +1)− l(l +1)− s(s +1)( )

J = 52

Espin−orbit =12hc !Α 5

2(52+1)− 3(3+1)− 1

2(12+1)⎛

⎝⎜⎞⎠⎟ = −2hc !Α

J = 72

Espin−orbit =12hc !Α 7

2(72+1)− 3(3+1)− 1

2(12+1)⎛

⎝⎜⎞⎠⎟ = + 3

2hc !Α

J = 52

J = 72

ΔEsplitting =

32hc !Α− −2hc !Α( ) = 72 hc !Α !Α : ~10 cm-1

Term Symbolsare used to specify angular momentum states of a multi-electron atom.

!L : orbital angular momentum

!S : spin angular momentum

!J : total angular momentum

2S+1LJground state Lithium (Li): 1s2 2s1

2S +1( ) multiplicity : one unpaired electron S = 12

2S +1= 2

angular momentum quantum number for atomic orbitals

symbol s p d f g hl 0 1 2 3 4 5

Total Orbital Angular Momentum Quantum Number

1 2

3

total orbital moment:

!L =

"Li

i=1

n

∑ !L1

!L2

!L3

longest added vector:

Lmax = L1 + L2 + L3shortest added vector:

Lmin = L1 − L2 − L3

Lithium atom:

1s2 2s 0,0,0{ } : L = 01s2 2p 0,0,1{ } : L = 1

1s2p3d : 0,1,2{ } Lmax = 2+1=3 or Lmin =2-1=1 L = 1 , 2 or 3

L = Lmin,Lmin +1,...,Lmax −1,Lmax{ } Clebsch-Gordan series

Roman Letters for Total Orbital Angular Momentum

symbol S P D F G HL 0 1 2 3 4 5

Lithium atom:

1s2 2s : L = 0 S 1s2 2p : L = 1 P 1s2p3d : 0,1,2{ } L = 1 , 2 or 3 P, D or F

What are the Roman letters for 2p3d ? 2p3d : {1,2} L = 1 , 2 or 3 P, D or F

What are the Roman letters for 2p4f ? 2p4 f : {1,3} L = 2 , 3 or 4 D, F or G

Total Angular Momentum Quantum Number

maximum magnitude of J: J max = L + S

minimum magnitude of J: J min = L − S

!J =!L +!S

Lithium atom:

1s2 2s : L = 0 S= 12

J= 12

1s2 2p : L = 1 S= 12

J= 32

, 12

⎧⎨⎩

⎫⎬⎭

2S1/2

2P3/2 ,2 P1/2

energy splitting

Self TestPlease write down the symbol terms for an excited carbon atom

electronic configuration: 1s2 2s2 2p13p1

(2S +1) multiplicity :

S = (12− 12, 12+ 12) = 0,1( )

total orbital momentum quantum number :

L = (1−1,1,1+1) = 0,1,2( ) (S,P,D)

total momentum quantum number :

J = L − S , L − S +1,...,L + S −1,L + S{ }

Continued

singlet state: S = 0 L = 0,1,2{ } J = 0,1,2{ }1S0,

1P1,1D2

triplet state: S = 1 L = 0,1,2{ } J = 0,1,2,3{ }

3S1S = 1 L = 0 J = 1 S = 1 L = 1 J = 0,1,2

3P0 ,3 P1, ,

3 P2S = 1 L = 2 J = 1,2,3

3D1,3D2,

3D3

In total, we have ten fine orbitals split by spin-orbit coupling

Selection Rule for Multi-Electron AtomsThe likeliness of electronic transitions:

ΔS = 0,ΔL = 0,±1,Δl = ±1,ΔJ = 0,±1, J = 0←×→ J = 0

ΔS : no spin flippingΔL = 0,±1: the change of total orbital quantum number can not be greater than 1

Δl = ±1: There must be a change on an individual electron's orbital shape

ΔJ = 0,±1 : the change of total angular momentum quantum number can not be greater than 1

J = 0 ←×→ J = 0 : a transition from J=0 to J=0 is forbidden

Ri→ f =

2π!

ψ i"r ψ f

2ρ f

Fermi’s Golden Rule:

transition dipole moment

Some Examples

3p 2P3/2 →2 S1/2 4s

ΔS = 2 − 2 = 0 ΔL = 0 −1= −1 ΔJ = 1/ 2 − 3 / 2 = −1Δl = 0 −1= −1

the transition is allowed

1D2 →3 D2

ΔS = 3−1= 2

the transition is forbidden due to spin-flipping

Review of Homework 6Review of Homework 109.26 Thallium, a neurotoxin, is the heaviest member of Group 13 of the periodic table and is found most usually in the +1 oxidation state. Aluminium, which causes anaemiaand dementia, is also a member of the group but its chemical properties are dominatedby the +3 oxidation state. Examine the issue by plotting the first, second and third ionizationenergies for the group 13 elements against atomic number.

kJ/mol I1 I2 I3

B (5) 800.6 2427

Al (13) 577.4 1816.6 2744.6

Ga (31) 578.8 1979 2963

In (49) 558.3 1820.6 2704

Tl (81) 589.3 1971.0 2878

I1< I2 < I3

Al : 3s2 3p1 +3

Tl : 6s2 5d10 6p1 +1Aufbau principle

Homework 11

Reading assignment: Chapters 9.8, 9.9, and 9.10

Homework assignment: Exercises 9.26 Problems 9.6

Homework assignments must be turned in by 5:00 PM, February 28th, Tuesday

to my mailbox in the Department Main Office located at Room 4000, Science and Engineering Hall