P460 - Spin 1

Spin and Magnetic Moments

• Orbital and intrinsic (spin) angular momentum

produce magnetic moments

• coupling between moments shift atomic energies

· Look first at orbital (think of current in a loop)

· the “g-factor” is 1 for orbital moments. The Bohr

magneton is introduced as natural unit and the “-”

sign is due to the electron’s charge

LgL

mvrLbutr

areacurrentAI

b

llmq

rqv

2

22

eb m

e

2

lblzlbll mgllg

llL

)1(

)1(2

P460 - Spin 2

Spin • Particles have an intrinsic angular momentum - called spin

though nothing is “spinning”

• probably a more fundamental quantity than mass

• integer spin --> Bosons half-integer--> Fermions

Spin particle postulated particle

0 pion Higgs, selectron

1/2 electron photino (neutralino)

1 photon

3/2 delta

2 graviton

• relativistic QM uses Klein-Gordon and Dirac equations for spin

0 and 1/2.

• Solve by substituting operators for E,p. The Dirac equation

ends up with magnetic moment terms and an extra degree of

freedom (the spin)

22222 :: mpEDmpEKG

P460 - Spin 3

Spin 1/2 expectation values • similar eigenvalues as orbital angular momentum (but SU(2))

• Dirac equation gives g-factor of 2

• non-diagonal components (x,y) aren’t zero. Just

indeterminate. Can sometimes use Pauli spin matrices to

make calculations easier

• with two eigenstates (eigenspinors)

200232.2

,,

||...||,)1(

21

212

432

23

212

21

22

sSg

s

z

z

g

SSSfor

ssSssS

bs

0

0

01

10

10

01

10

012224

32 2

i

iSSSS yxz

2

2

1

0

0

1

eigenvalueS

eigenvalueS

z

z

P460 - Spin 4

Zeeman Effect • Angular momentum->magnetic moment->energy shifts

• additional terms in S.E. do spin-orbit later. Right now

assume atom in external magnetic field

• look at ground state of H. L=0, S=1/2

dVEVwithVEE

shiftonperturbatiBE

nBnnnnn

B

*

2

21*

BgV

dVSdiagonalS

BSg

BSg

E

bSnn

zZ

Zbsbs

B

P460 - Spin 5

Spin 1/2 expectation values • Let’s assume state in a combination of spin-up and spin-down

states (it isn’t polarized).

• Can calculate some expectation values. Griffiths Ex. 4.2. Z-

component

• x-component

• as normalized, by inspection

1|||| 22

bawithb

aba

322

2

642

2622

2

61

)(

2

1

baS

bSaS

ilet

z

zz

361

2**

2

2

2**

)1(22)1()(

0

0

iiabba

b

abaSS x

tx

31

21

61

65

61

2

65

2

)()(

)(

x

x

Syprobabilit

Syprobabilit

P460 - Spin 6

• Griffiths Prob. 4.28. For the most general normalized spinor

find expectation values:

• just did x and z

• repeat for other

• note x and y component will have non-zero “width” for their

distributions as not diagonalized

222 ,,,,,, zyxzyx SSSSSSfindba

)(0

0

)(10

01

**2

2

2**

**2

**

abbab

abaS

bbaab

abaS

x

z

2

31222

4**

4**

42

**2

2

2**

222

)(01

10

01

10

)(0

0

SSSS

bbaab

abaS

baabb

abaS

zyx

x

ii

i

y

P460 - Spin 7

• Can look at the widths of spin terms if in a given eigenstate

• z picked as diagonal and so

• for off-diagonal

0)11()(

0

1

10

01

10

0101

4

222

442

2

22

zzz

z

SSS

S

Widths

0

1

4

222

442

2

2

2

22

)(

0

1

01

10

01

1001

00

1

0

001

xxx

x

x

SSS

S

S

P460 - Spin 8

• Assume in a given eigenstate

• the direction of the total spin can’t be in the same direction as the z-component

(also true for l>0)

• Example: external magnetic field. Added energy

puts electron in the +state. There is now a torque

which causes a precession about the “z-axis” (defined by the magnetic field)

with Larmor frequency of

Components, directions, precession

0

1

31

232

2

cos

S

S z BS

BE s

BSB bsgs

Bg bs

z

P460 - Spin 9

• Griffiths does a nice derivation of Larmor precession but at the 560 level

• to understand need to solve problem 4.30.

• Construct the matrix representing the component of spin angular

momentum along an arbitrary radial direction r. Find the eigenvalues and

eigenspinors.

• Put components into Pauli spin matrices

• and solve for its eigenvalues

Angles

kjir ˆcosˆsinsinˆcossinˆ

cossinsincossin

sinsincossincos

i

iSr

10|| ISSr

P460 - Spin 10

• Go ahead and solve for eigenspinors.

• Phi phase is arbitrary. gives

• if r in z,x,y-directions

kjir ˆcosˆsinsinˆcossinˆ

cossinsincossin

sinsincossincos

i

iSr

)tan(cos

sin

sin

)cos1(

)sin(cossincos

1

2sincos1

2

2

useeae

ab

aiba

b

aforS

ii

r

2

2

2

2

cos

sin1

sin

cos

i

ri

r efor

e

21

2

2

21

22

21

21

212

1

2

,,:

,0,:

1

0,

0

10:

i

iy

x

z

P460 - Spin 11

Combining Angular Momentum • If have two or more angular momentum, the

combination is also eigenstate(s) of angular momentum. Group theory gives the rules:

• representations of angular momentum have 2 quantum numbers:

• combining angular momentum A+B+C…gives new states G+H+I….each of which satisfies “2 quantum number and number of states” rules

• trivial example. Let J= total angular momentm

stateslllllm

l

12,1...1,

......,1,,0 23

21

221sinsin

,,0 21

21

21

gletdoubletglet

JJSLif

SSLLSLJ

z

ii

P460 - Spin 12

Combining Angular Momentum • Non-trivial example.

• Get maximum J by maximum of L+S. Then all possible combinations of J (going down by 1) to get to minimum value |L-S|

• number of states when combined equals number in each state “times” each other

• the final states will be combinations of initial states. The “coefficiants” (how they are made from the initial states) can be fairly easily determined using group theory (and step-up and step-down operaters). Called Clebsch-Gordon coefficients

2423

,,,

1,0,1,1

21

21

21

23

23

21

21

doubletquartetdoublettriplet

JJANDJJ

SLwithSLif

zz

zz

P460 - Spin 13

• Same example.

• Example of how states “add”:

• Note Clebsch-Gordon coefficients

23

23

21

21

23

21

21

21

23

21

21

21

23

21

21

21

23

21

21

23

23

21

21

1

0

1

1

0

1

JJJSL zzz

21

31

21

32

21

21

21

32

21

31

21

23

, 0 , 1

,

, 0 , 1

,

z z z z

z

z z z z

z

S L S L

J J

S L S L

J J

3

2,

3

1

2 terms

P460 - Spin 14

• Clebsch-Gordon coefficients for different J,L,S