Chapter7b–!ElectronSpinand!Spin4Orbit!Coupling!scienide2.uwaterloo.ca/~nooijen/Chem356/Chem+356+pdf/Ch_7b.pdf ·...

Winter 2013 Chem 356: Introductory Quantum Mechanics

96

Chapter 7b – Electron Spin and Spin-‐Orbit Coupling ................................................................................. 96

H-‐atom in a Magnetic Field: Electron Spin ............................................................................................ 96

Total Angular Momentum ................................................................................................................... 103

Chapter 7b – Electron Spin and Spin-‐Orbit Coupling

H-‐atom in a Magnetic Field: Electron Spin

If electron in orbital has angular momentum !L , one has a magnetic moment

!m = q

2m!L →

−e2me

!L

This magnetic moment can interact with a magnetic field and the interaction energy is given by

V = − !m ⋅!B

!B is measured in Tesla 1T = 1 Newton/(ampere meter)

If we take !B to be in z direction then

V = −Bzmz =

e2me

Bz Lz

And the total Hamiltonian would be

H = H0 +

e2me

Bz Lz

The wavefunctions we obtained ψ n,l ,m = fn,l (r)yl

m(θ ,ϕ ) are also eigenfunctions of H

including the magnetic field En,l ,m = En,l

(0) + e2me

B(m!)

each level ,n l splits in (2 1)l + sublevels

This is what would be expected, from classical considerations. This is not what is observed!


Chapter 7b – Electron Spin and Spin-‐Orbit Coupling 97

Stern and Gerlach passed silver atoms (with electron in s-‐orbital) through inhomogeneous magnetic field.

1922! They found this splits the beam into two. Classically one would expect (2 1)l + lines for particle with

angular momentum l . Here 2 1 2l + = → 12

l = !!

This was the first evidence that electrons need another quantum number: half-‐integer angular momentum. Jumping to what we know now, we introduce an intrinsic angular momentum operator

Sx , S y , Sz

S 2 = Sx2 + S y

2 + Sz2

This operator is postulated to have exactly the same commutation relations as ˆ ˆ ˆ, ,x y zL L L

Hence

Sx , S y⎡⎣ ⎤⎦ = i!Sz

Sz , Sx⎡⎣ ⎤⎦ = i!S y

S y , Sz⎡⎣ ⎤⎦ = i!Sx

From out general discussion we know the possible eigenstates as

S2 s,ms = !2s(s+1) s,ms

Sz s,ms = ms! s,ms

Where I indicate the states as s,ms : Dirac bracket notation (page 159. MQ)

The splitting of silver atom beam indicates eigenstates

1 1,2 2ss m α= = =

1 1,2 2ss m β= = − =



Everything we discussed on operators and angular momentum applies to spin, also the states α , βare orthonormal

α |α = α *(σ )α (σ )dσ = 1∫

β |β = β *(σ )β(σ )dσ = 1∫

α |β = α *(σ )β(σ )dσ = 0∫

What are the ‘coordinates’ that we integrate over? We do not have a clue!! But, also, we do not need it:

We know α , β are orthogonal because they are eigenstates of a Hermitian operator ˆzL

with different eigenvalues ± 1

2! . Hence they must be orthogonal! (See chapter 4).

We now have 2 sets of angular momentum operators

ˆ ˆ ˆ, ,x y zL L L : orbital angular momentum

ˆ ˆ ˆ, ,x y zS S S : spin angular momentum

ˆˆ , 0L Sα β⎡ ⎤ =⎣ ⎦

The operators 2 2ˆ ˆˆ ˆ, , ,z zL S L S all commute, and they also commute with the Hamiltonian

0H T V= + , for the H-‐atom

Then we can characterize each eigenstate through

, , , ,l sn l m S m 5 quantum numbers

Let us facilitate the notation somewhat Note: Later on this will require modification, as we are neglecting relativistic effects, which introduce so-‐called spin-‐orbit coupling.



We can indicate the l -‐quantum number as s, p, d and get functions

12P , 2P0 , 12P− 1,0, 1lm = −

We could then write, including spin 12Pα , 02Pα , 12P α−

12Pβ , 02P β , 12P β−

In what follows we will focus on one particular n quantum number, which we can suppress. Moreover I can indicate the β -‐spin function by an overbar.

Then we get the 6 p-‐functions

p1 , p0 , p−1 , p1 , p0 , p−1

Or the d-‐functions

2d , 1d , 0d , 1d− , 2d− , 2d , 1d , 0d , 1d− , 2d−

This indicates the , , ,s sl m S m quantum numbers

In this way we can label the exact eigenstates of the (non-‐relativistic) Hamiltonian. The splitting of the lines in a magnetic field would then be determined by the Hamiltonian

H = H0 +

e2me

Bz Lz +ge

2me

BzSz where 2.0011...g ≈

This factor g determining the ratio between spin and orbital interactions with the magnetic field, can

be calculated using relativistic quantum field theory. (Schwinger, Tomanaga, Feynman). Far beyond our aim. It agrees to about 10 digits with the experimental value! (that is like measuring the distance from here to New York up to a millimeter!)

The p1 , p1 functions are eigenfunctions of this magnetic Hamiltonian, and we can easily calculate the

energy splitting. Unfortunately, this does not give correct results!! The splitting due to the magnetic field is very small. There are other corrections to the energy levels in Hydrogen atom due to relativity (think Einstein). They are of at least comparable importance, and cannot be neglected when discussing magnetic effects. The relativistic Hydrogen atom is described by the Dirac equation. This is far more complicated than we wish to discuss.



One can approximate effects by including so-‐called spin-‐orbit interaction in the Hamiltonian.

H0( R) = T + V + ξ(r)L ⋅ S

L ⋅ S = LxSx + LyS y + LzSz

It is called L S⋅ coupling or spin-‐orbit coupling Then the magnetic interaction can be included as

( )0

ˆˆ ˆ2 2

Rz z

e e

e geH H L Sm m

= + +

Let us first examine the energy levels for the non-‐relativistic Hamiltonian, i.e. we neglect the spin-‐orbit or L S⋅ coupling term.

0ˆˆ ˆ ˆ

2 2z z z ze e

e eH H B L g B Sm m

= + +

Let us denote !

e2me

Bz

⎛⎝⎜

⎞⎠⎟= γ and use 2g = , then

H = H0 +

γ!

Lz +2γ!

Sz

Let us consider the allowed 2 1p s+ emission lines: Normal Zeeman effect:

0H , only include zL include zS



Due to ˆzS all α -‐levels go up by γ all β -‐levels go down by γ . Since transitions cannot change spin, 0smΔ = , I get same transitions as without spin!!

Our conclusion thus far.

If one does not consider spin levels split in a magnetic field using ˆ2 z

e

e Lm

p→ 3 equal spaced levels

d → 5 equal spaced levels If we include spin, for single electron states then all α -‐states shift up by 1 unit of γ ! , all β -‐levels shift

down by 1 unit of γ ! , and the transition energies 'np n sα α→ are not affected by spin. Moreover all

multiples are split by the same amount

e2me

!Bz . We would not see the effects of spin.

This is what was originally observed in early experiments. It is called the ‘normal Zeeman effect’ and it was explained by Lorentz (two Dutch physicists). It appears as if only the Lz term is present. However we do observe the effects of spin in emission spectra! The story is more complicated. The complications occur already for the H-‐atom without a magnetic field. There is a substantial correction due to what is called spin-‐orbit interaction. A good way to think about this is as follows: We usually think of the electron as wizzing around the nucleus. From the point of view of the electron we can just as easily think that the nucleus is wizzing about the electron. This moving nucleus, with its angular momentum generates a magnetic field. This magnetic field interacts with the spin of the electron.

Compare the electron with your position on the spinning earth. The sun rises and sets from our standing still point of view, and moves with incredible velocities, in this frame. For a charged particle the magnetic force would be large. It is called spin-‐orbit coupling



2

2 2 3

1 ˆˆ2so

e

eH L Sm e r

= ⋅

Spin-‐orbit coupling is usually said to be a relativistic effect. This is because it arises in a natural way from the fully relativistic Dirac equation. So does spin; it arises naturally. And so do particles and antiparticles, which also arise from the Dirac equation. Let me say something more about spin. The spin operators are best represented by matrices.

σ x = !

0 11 0

⎛

⎝⎜⎞

⎠⎟,

σ y = !

0 −ii 0

⎛

⎝⎜⎞

⎠⎟,

σ z = !

1 00 −1

⎛

⎝⎜⎞

⎠⎟

These matrices satisfy the commutation relations of angular momentum

σ x ,σ y⎡⎣ ⎤⎦ = i!σ z

Moreover

Sz

10

⎛

⎝⎜⎞

⎠⎟= ! 1 0

0 −1⎛

⎝⎜⎞

⎠⎟10

⎛

⎝⎜⎞

⎠⎟= ! 1

0⎛

⎝⎜⎞

⎠⎟

Sz

01

⎛

⎝⎜⎞

⎠⎟= ! 1 0

0 −1⎛

⎝⎜⎞

⎠⎟01

⎛

⎝⎜⎞

⎠⎟= −! 0

1⎛

⎝⎜⎞

⎠⎟

Hence α = 1

0⎛

⎝⎜⎞

⎠⎟

01

β ⎛ ⎞= ⎜ ⎟⎝ ⎠

2*2 matrices have only 2 eigenvectors. This is why we have only α , β

The Dirac equation is a 4*4 matrix equation and we get (2 spin * 2 mass) solutions. The splitting due to the magnetic field is very small. There are other corrections to the energy levels in Hydrogen atom due to relativity. They are of at least comparable importance, and cannot be neglected when discussing magnetic effects. The relativistic Hydrogen atom is described by the Dirac equation. This is far more complicated than we wish to discuss. One can approximate the effects by including spin-‐orbit interaction in the Hamiltonian.

( )0

ˆˆ ˆ ˆ ˆ( )RH T V r L Sξ= + + ⋅

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆx x y y z zL S L S L S L S⋅ = + +

It is called L S⋅ coupling or spin-‐orbit coupling Then the magnetic interaction can be included as



( )0

ˆˆ ˆ2 2

Rz z

e e

e eH H L g Sm m

= + +

Our original spin-‐orbitals p1, p0 , p−1, p1, p0 , p−1 are however not eigenstates of the ( )0ˆ RH including L S⋅

coupling

We can classify the eigenstates of ( )ˆ RH by doing a little more angular momentum theory. Let me sketch the result, as this is, finally, an accurate description.

Total Angular Momentum

We have ˆ ˆ ˆ, ,x y zL L L and ˆ ˆ ˆ, ,x y zS S S

Moreover

Lα , Sβ⎡⎣ ⎤⎦ = 0 , , ,x y zα β =

This is because L and S act on different coordinates Now we define total angular momentum

ˆˆ ˆx x xJ L S= + ; ˆˆ ˆ

y y yJ L S= + ; ˆˆ ˆz z zJ L S= +

J 2 = Jx

2 + J y2 + Jz

2

ˆ ˆ,x yJ J and ˆzJ satisfy the usual commutation relations:

ˆ ˆˆ ˆ ˆ ˆ, ,x y x x y yJ J L S L S⎡ ⎤⎡ ⎤ = + +⎣ ⎦ ⎣ ⎦

ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ, , , ,x y x y x y x yL L L S S L S S⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤= + + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

= i!Lz + 0+ 0+ i!Sz

= i!J z

Moreover

J 2 = Jx

2 + J y2 + Jz

2

2 2ˆˆ 2L S L S= + + ⋅ x x y y z zL S L S L S L S⋅ = + +

( )2 2 212J L S= − −

And

2 2 2, , , 0x x xJ L L L S L⎡ ⎤ ⎡ ⎤ ⎡ ⎤= + =⎣ ⎦ ⎣ ⎦ ⎣ ⎦

Likewise



2 2 2ˆ ˆˆ , , , 0J S L S S Sα α α⎡ ⎤ ⎡ ⎤⎡ ⎤= + =⎣ ⎦⎣ ⎦ ⎣ ⎦

But then also

[ ]2 2 2ˆˆ ˆ, , 0J J S L J L Sα α⎡ ⎤− − = ⋅ =⎣ ⎦

And 2 2 2 2, , 0J L J S⎡ ⎤ ⎡ ⎤= =⎣ ⎦ ⎣ ⎦

Hence we can derive without too much trouble that the operators 2 2 2ˆˆ ˆ, ,J L S and ˆzJ all commute.

Moreover these operators commute with L S⋅ , and also with ˆˆ( )r L Sξ ⋅ .

It then follows that the angular momentum operators 2 2 2ˆˆ ˆ, ,J L S , ˆzJ commute with the relativistic

Hamiltonian ( )ˆ RH . And we can classify the states with quantum numbers

, , , , jn l s j m

As for the non-‐relativistic case, the angular momentum problem, defining , , , jl s j m is independent

from the radial equation, and can be solved once and for all.

2 , , , jL l s j m = l(l +1)!2 l,s, j,mj

2ˆ , , , jS l s j m = s(s+1)!2 l,s, j,mj

2ˆ , , , jJ l s j m = j( j +1)!2 l,s, j,mj

ˆ , , ,z jJ l s j m = mj!

2 l,s, j,mj

These equations hold for the Hydrogen atom, but later on we will see that they are very similar for many-‐electron atoms, which also have spherical symmetry.

Can we deduce what the eigenstates of 2J and ˆzJ might be?

First we note that eigenfunctions of zJ are easy.

J z l,s,ml ,mj = ml + ms( )! l,s,ml ,mj

Eg. J z p1 = 1+ 1

2⎛⎝⎜

⎞⎠⎟!p1

J z p0 = 0− 1

2⎛⎝⎜

⎞⎠⎟!p0

We also know that acting with J+ on ,j j should give 0: the highest jm value in the multiplet.

It is easy to find the highest function: lm l= , sm s=



Hence in the p-‐manifold p1 = p1α mj =

32,

32

j = is the highest jm function.

In the d-‐manifold it is 2 2d d α= with 1 522 2jm = + =

Acting with J− we can get the other function in the multiplet.

We get (2 1)j = in 32 1 42⋅ + = functions

We are left with 2 functions in the p-‐manifold (6 spin-‐orbital in total). This creates a 12

j = multiplet.

1 1,2 2

,

12

,− 12

2 functions.

How does this work, for the d-‐ functions? 2l = highest ml .

= 6 functions

In total we have 10 functions other multiplet has 4 functions

For the Hydrogen atom we can construct always

1 522 2j j jm l s= + = + = 5

2j→ =

5 3 1 1 3 5, , , , ,2 2 2 2 2 2jm = − − −

( )2 1J +→

→ 12

j l= − 1 322 2

= − =

12

j l⎛ ⎞= +⎜ ⎟⎝ ⎠12 1 2 22

l l⎛ ⎞+ + = +⎜ ⎟⎝ ⎠



is the total number of spin-‐orbitals of angular momentum .

We label the final eigenstates as 2S+1 LJ

The j-‐multiplet have slightly different energies due to the coupling, which for

equal is seen to depend on the quantum number.

Finally, we are ready to discuss how these relativistic levels split in a magnetic field; Within a multiplet the levels split in equally spaced levels due to a magnetic field

The splitting depends on

Allowed transitions: , ,

12

j l⎛ ⎞= −⎜ ⎟⎝ ⎠12 1 22

l l⎛ ⎞− + =⎜ ⎟⎝ ⎠( )2 2 1l⋅ +

( )2 2 1l⋅ + l

( )2 2 22L S J L S⋅ = − −

,L S J

J 2 1J +

( ), ,n l j

1lΔ = ± 0sΔ = 0, 1JmΔ = ±



All different energies. Tiny splitting, but this is how we can experimentally access degeneracies of energy levels. The splitting of energy levels in a magnetic field is a complicated subject, because relativistic effects have to be considered at the same time. Magnetic transitions using nuclei as in NMR are much simpler as we only need to consider angular momentum theory itself. Better picture of transitions including spin-‐orbit:

10 different transitions, 10 different frequencies

Selection rules

, , ,

No transitions

2 2 p32

→2 2s12

(mj =12

) 3 transitions

2 2 p12

→2 2s12

(mj =12

) 2 transitions

2 2 p32

→2 2s12

(mj = − 12

) 3 transitions

0B =

1lΔ = ± 0sΔ = 0, 1jΔ = ± 0, 1jmΔ = ±0 0J J= → =

0, 1jmΔ = ±



2 2 p12

→2 2s12

(mj = − 12

) 2 transitions

10 transitions in total

To good approximation one can evaluate the shifts in magnetic field from

Neglecting spin-‐orbit coupling.

2 ˆˆ2 2z z

e e

e eL Sm m

+

Chapter7b–!ElectronSpinand!Spin4Orbit!Coupling!scienide2.uwaterloo.ca/~nooijen/Chem356/Chem+356+pdf/Ch_7b.pdf ·...

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