This interaction is weaker than the Coulomb interaction and is

responsible for the fine-structure of atomic lines.

How do we understand the spin-orbit interaction?

• The electron has a spin s, as well as orbital angular momentum l.

• For each angular momentum, there is an associated

magnetic moment, and

• Spin-orbit = interaction between and

Let’s start with the definition of magnetic moment….

lµ

sµ

Spin-orbit interaction (or “coupling”) H&W Chapter 12

lµ

sµ

electron orbit = current loop

the magnetic moment is defined as where I is the current

and A the area of the orbit.

is a vector perpendicular to the plane of the orbit.

Now let’s show that is linked to the orbital angular momentum:

π

ω

ω

π

2

2T

e

T

eI

−=

−==

Magnetic moment lµ

22

2

1

2rer

eAIl ωπ

π

ωµ −=

−=⋅=

AIl ⋅=µ

lµ

lµ

orbit

period

2

00 rmvrmprl ω==×=rrr

orbital

ang. mom.l

m

el

rr

02−=µ

r

v

+

e

lµr

lr

A

o

Bm

e

2

h=µ Bohr magneton

h=l

Note that and point in opposite directions because the electron is

negatively charged.

Let’s consider the case : this is the orbital angular

momentum of the ground state in Bohr’s theory, and is therefore

taken as the unit of angular momentum in atoms.

The modulus of the corresponding magnetic moment is known as

Bohr magneton:

It is unit of magnetic moments in atoms.

lr

lµr

)1()1(2

+=+= llllm

eB

o

l µµ hr

More generally, for a state with quantum numbers l and :

hlZ ml = lml ±±±= ....2,1,0with magnetic quantum number

lBz

o

Zl mlm

eµµ −=−=

2,Component along z

Modulus

lm

)1( += lll hr

We also can apply these definitions to the electron spin….

Electron spin

)2

1(with

±=

=

s

sZ

m

ms h

hhr

2

3)1( =+= sss

• An intrinsic property like charge or mass.

• Its existence proven by Stern-Gerlach experiment (see later), and

justified by Dirac’s relativistic quantum theory.

• It’s an intrinsic angular momentum with quantum number2

1=s

Hence we can apply the same theory that we saw for the orbital

angular momentum (see Hydrogen atom- angular momentum aside).

Component along z

Modulus

Spin magnetic moment sm

eg

o

ss

rr

2−=µ

0023.2=sgWhere is a numerical factor known as the Lande factor

(or “g factor”). Classically it should be 1, but is 2 according to

Dirac’s relativistic quantum theory.

( for orbital angular momentum)1=lg

Next, let’s consider the magnetic interaction of a magnetic

moment with a magnetic field :

Brr

×= µτtorque

(Note that the potential energy is minimised if magnetic moment is

parallel to B, i.e. α=0.)

Magnetic potential energy:

µr

Br

αµr

Br

BB

BdBdBVmag

rr

rr

⋅−=−=

=−==×= ∫∫

µαµ

αµααµαµ απ

α

π

α

π

cos

][cossin)( 2/

2/2/

BVmag

rr⋅−= µ

lssl BVrr

⋅−= µ,

lBr

lBr

s

s

Spin-orbit interaction

r

-rlBr

Orbiting proton

produces a field

at the site of

the electron.lB

The spin, hence , has two possible orientations with respect

to the field: “parallel”, with lower energy, and “anti-parallel” with

higher energy

we expect energy levels in atoms to split in two (fine structure).

sµr

Note that this is not the only magnetic interaction that is taking place in an atom. The nucleus

also has a spin, hence a magnetic moment. The nuclear magnetic moment interacts with the

magnetic field produced by the orbiting electron. This interaction is much weaker than the

spin-orbit interaction considered here, and is responsible for the hyperfine structure of the

energy levels – see later.

Biot-Savart law: ][4

)]([4 3

0

3

0 rvr

Zerv

r

ZeBl

rrrrr×−=−×=

π

µ

π

µ

rvml

vmrlrrr

rrr

×=−

×=

0

0l

mr

ZeBl

rr

0

3

0

4π

µ=

Our goal is to determine an expression for the fine structure starting

from . Let’s start by writing the B-field in terms of the

orbital angular momentum: lssl BVrr

⋅−= µ,

The – sign here is because we are

in the electron’s frame of reference,

see previous slide

lmr

ZeBl

rr

0

3

0

42

1

π

µ×=

from full relativistic

calculation

(“Thomas factor”)

sm

es

m

eg

oo

ss

rrr−≈−=

2µ

)2( ≈sg

)(8

)(32

0

0

2

0

, lsrm

ZeBs

m

eV lsl

rrrr⋅=⋅=

π

µ

sla

V sl

rr

h⋅=

2,32

0

2

0

2

8 rm

Zea

π

µ h=

We define the spin-orbit coupling constant a:

( )

12

13

4

+

+

∝

llln

Za

(With this definition a has the dimension of an energy.)

The constant a is measurable from spectra. We won’t do the

calculation, but it can be shown that

dVrr

ψψ∫=3

*

3

11

n valueexpectatio from

Heavier atoms larger spin-orbit coupling

Larger n smaller spin-orbit coupling

222

222

2,

ˆ ˆ ˆ

)(2

slj

slja

V sl

rrr

h−−=

… let’s now consider the scalar product .

It is possible to express this in terms of the total angular momentum:

),cos( slslslrrrr

=⋅

sljrrr

+= jr

lr

sr

),cos(2222

slslsljrrrrr

++=

(These vectors are operators

in quantum mechanics)

jjjm j −−= ,..,1,

modulus^2: with

projection along z-axis: with

jr

Properties of :

• It is an angular momentum, hence we can apply the results

we saw in the “angular momentum aside”.

new quantum numbers j and such that:jm

Ψ+=Ψ )1(ˆ 22jjj h

Ψ=Ψ jz mj hˆ

... ,2

3 1, ,

2

1

2

integer ==j

[ ] [ ] 0ˆ,ˆ , 0ˆ,ˆ 2222 == sjlj• We have the commutation relations

[ ] [ ] ),, where0ˆ,ˆˆ,ˆ from follow(which 22 zyxisjlj ii ===

hence we can choose a set of common eigenfunctions of

, and calculate on these eigenfunctions….222 ˆ,ˆ,ˆ jsl slV ,

2

1±=±= lslj

[ ])1()1()1(2

)(2

222

2, +−+−+=−−= sslljja

slja

V sl

rrr

h

… and so we can replace the operators with the corresponding

eigenvalues:

[ ])1()1()1(2

, +−+−+= sslljja

V sl

This is our final result for spin-orbit coupling:

shift in energy due

to spin-orbit coupling

But we still haven’t discussed how to calculate j….

From the theory of coupling of angular momenta, it turns out that

e.g. for a p state (l=1)j=1/2

j=3/2

Example:

fine structure of sodium atoms

• state 3s (l=0)

there is no fine structure

(physically, there is no angular momentum,

hence no “current loop” and )

• state 3p

0 and 2/1 , == slVj

0=lB

j=3/2 and

j=1/2 and aa

V sl −=

⋅−−⋅=2

3

2

12

2

3

2

1

2,

22

3

2

12

2

5

2

3

2,

aaV sl =

⋅−−⋅=

fine structure for 3p a/2

aa

2

3j=3/2

j=1/2

3s 3p transition is split into a doublet

(D lines: D1 and D2)

Sodium doublet viewed in a Fabry-

Perot interferometer:

Nomenclature for naming energy terms:

j

sln

12 +

orbital angular momentum

from earlier =s,p,d,f etc...

total angular

momentum

spin term

principal

quantum

number

e.g. ground state of sodium is

(the spin term is sometimes omitted, like in the previous slide, because

it’s the same for all the levels.)

2/1

23 s

Let’s see how the spectral lines split

due to fine-structure:

• series 3s np: pairs of lines

(all p states are double)

• series 3p nd: triple lines

This is because:

d state (l=2)j=5/2

j=3/2

and furthermore we have a new selection rule on j

hence only these three transitions are allowed:0,1±=∆j

2/13p

2/33p

2/5n d

2/3n d

Summary for FINE STRUCTURE

• spin-orbit coupling

• all energy states (except s-states) are split into dublets

• split (larger split in heavy atoms)4

Z∝

sla

BV lssl

rr

h

rr⋅=⋅−=

2, µ

But this is not the end with fine structure….

} 48476 FSE

slrel

nRy

n EEEE

structure fine

,

/

structuregross

2

++=

−

relativistic

mass change

spin-orbit coupling

…Hydrogen is a special case:

Small spin-orbit comparable to relativistic corrections

(remember Bohr-Sommerfeld:

in that model the fine structure was due to relativistic effects)

In heavier atoms, spin-orbit coupling is stronger and relativistic

effects can be neglected.

For the H-atom:

)(8 32

0

0

2

lsrm

e rr⋅

π

µ

−

+−=

njn

EE n

FS4

3

2/1

12α

Result from Dirac’s relativistic theory:

where the dimensionless quantity

is the fine structure constant.

(You can verify that α=v/c with v= velocity of 1st Bohr orbit.)

137

1

2 0

2

==hc

e

εα

Bohr Dirac QED

Example: let’s apply this result to the n=2 level of Hydrogen.

Here we have an s state and a p state:

2/3 ,2/11

2/10

=⇒=

=⇒=

jl

jl Can calculate

for these values of jFSE

(relativistic QM)

l=1, j=3/2

n=2

GHz10≈

l=0,1, j=1/2 MHz1057 2/1

22 p

2/1

22 s

2/3

22 p

This is a further splitting that

can only be explained by

Quantum Electrodynamics (QED)…

Significance of Lamb shift

In QED the electromagnetic field is quantised.

A quantised field has a zero point energy (analogous to the ground

state of the simple harmonic oscillator). This means that even in the

case of number of photons=0 (“vacuum”), there is a fluctuating

electric field.

Due to this fluctuating field the electron performs small oscillations

and its charge is effectively smeared out. It can be shown that the

Coulomb potential acting on the electron is different than that for a

point charge. This difference causes the Lamb shift.

(Note: this fluctuating electric field is also responsible for

spontaneous emission.)

… this is known as the Lamb shift of the s level.

It was first observed in 1947-1952 and was crucial

for the development of QED.