P460 - Spin 1

Spin and Magnetic Moments(skip sect. 10-3)

• 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 the 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)).

No 3D “function”

• 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

zkijkji

g

SSSfor

ssSssSSSS

bs

0

0

01

10

10

01

10

01

2

222

432 2

i

iSSS

SS

yxz

ii

2

2

1

0

0

1

eigenvalueS

eigenvalueS

z

z

P460 - Spin 4

Spin 1/2 expectation values • “total” spin direction not aligned with any component.

• can get angle of spin with a component

3

1

43

21

cos

S

Sz

P460 - Spin 5

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

spin-down states.

• expectation values. z-component

• x-component

• y-component

1|||| 22

bawithb

aba

)(

10

01||

222

2**

ba

b

abaSS zz

)(0

0 **2

2

2** abbab

abaSx

)(0

0 **2

2

2** abbaib

a

i

ibaS y

P460 - Spin 6

Spin 1/2 expectation values example

• assume wavefunction is

• expectation values. z-component

• x-component

• Can also ask what is the probability to have different

components. As normalized, by inspection

• or could rotate wavefunction to basis where x is diagonal

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

322

2

642

2622

2

)(

baS

bSaS

z

zz

2

16

1i

P460 - Spin 7

• Can also determine

• and widths

2

31222

4**

4**

42

4**

4**

42

4**

4**

42

222

222

222

)(10

01

10

01

)(0

0

0

0

)(01

10

01

10

SSSS

bbaab

abaS

bbaab

a

i

i

i

ibaS

bbaab

abaS

zyx

z

y

x

))(1(4

)(

))(1(4

)(

))(1(4

)(

2**2

222

2**2

222

2**2

222

bbaaSSS

abbaSSS

abbaSSS

zzz

yyy

xxx

P460 - Spin 8

• 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- example

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 9

• 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 10

• Hamiltonian for an electron in a magnetic field

• assume solution of form

• If B direction defines z-axis have Scr.eq.

• And can get eigenvalues and eigenfunctions

Precession - details

ti

ti

be

aet

b

a

m

egB

m

egB

)()0(

1

0

40

1

4

Bm

egH

4

2S

)(

)(

t

t

dt

diH

10

01BB

10

01

4m

Beg

dt

di

P460 - Spin 11

• Assume at t=0 in the + eigenstate of Sx

• Solve for the x and y expectation values. See how they precess around

the z-axis

Precession - details

ti

ti

e

et

b

a

b

a

b

a

2

1)(

1

1

2

1

201

10

2

ti

eebaab

iS

tee

abbaS

titi

y

titi

x

2sin2

)2

(2

)(2

2cos2

)2

(2

)(2

22**

22**

P460 - Spin 12

• can look at any direction (p 160 or see Griffiths 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

Arbitrary Angles

kjir ˆcosˆsinsinˆcossinˆ

cossinsincossin

sinsincossincos

i

iSr

10|| ISSr

P460 - Spin 13

• 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 14

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

combination is also an 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 momentum

stateslllllm

l

12,1...1,

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

21

221sin

,,0 21

21

21

doubletdoubletglet

JJSLif

SSLLSLJ

z

ii

kijkji SiSS ,

P460 - Spin 15

Combining Angular Momentum • Non-trivial examples. add 2 spins

• add spin and orbital angular momentum

1322

sin

0,01,0,1,1

,

21

21

221

121

221

1

glettripletdoubletdoublet

JJANDJJ

SSJ

SSwithSSif

zz

zz

2423

,,,

1,0,1,1

21

21

21

23

23

21

21

doubletquartetdoublettriplet

JJANDJJ

SLwithSLif

zz

zz

P460 - Spin 16

Combining Angular Momentum • 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 “coefficients” (how they are made from the initial states) can be fairly easily determined using group theory (step-down operaters). Called Clebsch-Gordon coefficients

• these give the “dot product” or rotation between the total and the individual terms.

mlmlmm

mlmlmm

mmmmm

mmmmtotalml

mmmmtotalml

21

21

2121

2121

2121

P460 - Spin 17

Combining Angular Momentum • example 2 spin 1/2• have 4 states with eigenvalues 1,0,0,-1. Two 0 states

mix to form eigenstates of S2

• step down from ++ state

• Clebsch-Gordon coefficients

1

00

1

,,,

z

zz

z

S

SS

S

)1)((

21

21

mlmlC

SSS

SSS zzz

orthogonalml

ml

mlmlCmlS

S

CS

CS

)(2

10,0

)(2

10,1

0,120,1)1,1(1,1

)2

(2

),(

),(

21

21

2

21

21

1

2

1

P460 - Spin 18

Combining Ang. Momentum • check that eigenstates have right eigenvalue for S2

• first write down

• and then look at terms

• putting it all together see eigenstates

21212122

21

21212122

21

2122

21

221

2

2

222

2)(

SSSSSSSS

SSSSSSSS

SSSSSSS

zz

yyxxzz

yx iSSS

)(2

1X

XXSSSS

SSSSand

SSSSwith

XXSS

XXS

XSSXS

zz

22112

212

221

1212

21

222

221

21

21

)(

0

0

)2

)(2

(22

4

3

4

3))()((

2

1

XXXS

0

2)1( 2

21

43

4322

P460 - Spin 19

• L=1 + S=1/2

• 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

3

2,

3

1

2 terms

21

31

21

32

21

21

21

32

21

31

21

23

01

01

mj

mj

SL zz

P460 - Spin 20

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