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)
  I  A  current  area
2
 2qv
but L  mvr
r  r

q
2m

L  l   gl
b

b 

L
e
2me
· 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
L
  l (l  1) 
2
l
 g l b
l (l  1)
P460 - Spin
 zl
  g l b ml
1
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
D
2
graviton
•
relativistic QM 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)
KG : E 2  p 2  m 2 D : E 
P460 - Spin
p2  m2
2
Spin 1/2 expectation values
• similar eigenvalues as orbital angular momentum (but SU(2)). No 3D
“function”
S , S   
i
j
ijk
S k 
S 2  s ( s  1) 2 , S z   | s | ... | s |
for S 
S2


s  
1
2
1 3
2 2

2 

g s b S

3
4
2 , Sz

1
2
,
1
2

g s  2.00232  2
• Dirac equation gives g-factor of 2
P460 - Spin
3
Spin 1/2 expectation values
• non-diagonal components (x,y) aren’t zero. Just indeterminate. Can
sometimes use Pauli spin matrices to make calculations easier

Si 
i  S2 
2
1

0

0
 
2 
1

Sz 
Sx

2
3 2
4
1

0

0

1

0 

 1

1

0

Sy 

2
0

i

 i

0 

• with two eigenstates (eigenspinors)
1
 
  
 0
  S z eigenv alue 
 
 
  
1
  S z eigenv alue 
 
 0
P460 - Spin

2

2
4
Spin 1/2 expectation values
• “total” spin direction not aligned with any component.
• can get angle of spin with a component
cos  
Sz

S

1

2

3

4
P460 - Spin
1
3
5
Spin 1/2 expectation values
• Let’s assume state in an arbitrary combination of spin-up and spindown states.
a
 
  a   b     with | a |2  | b |2  1
b
• expectation values. z-component
1
b*  2 
0

S z   | S z |   a *


2
0  a 





 1 b 

(a 2  b2 )
• x-component
• y-component
S x  a
S y  a
*
0
b 

2
*
*
0
b  
2i
*
P460 - Spin
 a 

b

0
 

2

2
( a*b  b*a )
 2 i  a 
   i 2 (a*b  b*a )
0  b 
6
Spin 1/2 expectation values example
 
• assume wavefunction is
1
6
1  i 

 2 



• expectation values. z-component
Sz 
Sz
• x-component

2

 a2 

2
Sz  
(a 2  b2 )  
Sx

2
6

2
 b2 
4
6

3

  Sx   a
t

2
( a *b  b * a ) 
*
 1
2 6

0
2  a 

b 
b



0
2
 
(1  i ) 2  2(1  i ) 
*


3
• Can also ask what is the probability to have different components. As
normalized, by inspection probabilit y(S x  2 )  56
probabilit y( S x   2 )  16  ( 56  16 ) 12  13
• or could rotate wavefunction to basis where x is diagonal
P460 - Spin
7
•
Can also determine
S x2 
2
4
a
*
S y2 
2
4
a
*
S z2 
2
4
a
*
0
b* 
1

0
b* 
i

1  0


0
 1
 i  0


0 
 i
 i  a 

b

0 
 
2
4
( a *a  b*b) 
2
4
1
b* 
0

0  1


 1
 0
0  a 

b

 1
 
2
4
( a *a  b*b) 
2
4
S x2  S y2  S z2 
1
3
1  a 

b

0
 
2
4
( a *a  b*b) 
2
4
S2
• and widths
( DS x )  S
2
2
x
( DS y ) 2  S y2
( DS z ) 2  S z2
2
 Sx

(1  ( a *b  b*a ) 2 )
4
2
2
 Sy

(1  ( a *b  b*a ) 2 )
4
2
2
 Sz

(1  ( a *a  b*b) 2 )
4
2
P460 - Spin
8
Widths- example
• Can look at the widths of spin terms if in a given eigenstate
 1
  
 0

 
• z picked as diagonal and so
S
2
z

2

4
1
1
0 
0

( DS z ) 2  S z2  S z
0  1


 1
 0
2

2
4
• for off-diagonal
Sx
 1
S x2

2
4
( DS x ) 2 
0  1 
2

 0
 4
 1
 
(1  1)  0
0
0 

2
1
S x2
 1 

 0
0
0
 
1  0 1  1 
0
2
0 


1

1

 0
 4
0
0


 
2
2
 Sx

4

2
P460 - Spin
9
Components, directions, precession

• Assume in a given eigenstate
 1

 0

 
• the direction of the total spin can’t be in the same direction as the zcomponent (also true for l>0)
Sz 
S

2
S2 
3
2
 cos  
z
1
3
B
• Example: external magnetic field. Added energy

DE   s  B
puts electron in the +state. There is now a torque
 



g s b
  s  B    S  B
which causes a precession about the “z-axis” (defined by the magnetic
g s b B
field) with Larmor frequency of
 

P460 - Spin
10
Precession - details
• Hamiltonian for an electron in a magnetic field

eg 
H 
 B
4m
• assume solution of form
   (t ) 
 
  (t ) 

 


 
S 

2
H  i
d
dt
• If B direction defines z-axis have Scr.eq.

1
  B  B
0


0 

 1

i
d
dt



 

egB



4m

• And can get eigenvalues and eigenfunctions
1

0

0   


 1
  




1
 0
egB
egB








 0
1

4m
4
m
 
 
 ae i t 
a
 ( 0)  
b
  (t )  
 be i t 

 


 
P460 - Spin
11
Precession - details
• Assume at t=0 in the + eigenstate of Sx

2
0

1

1  a 
a
 a








b
b
b
 
0
2
 
 
 
  (t ) 
1
2
 e i t

 ei t

1
2
 1

 1

 




• Solve for the x and y expectation values. See how they precess around
the z-axis
Sx
Sy

 e 2 i t  e 2 i t

*
*

(a b  b a ) 
(
) 
cos 2 t
2
2
2
2
i
 e 2 i t  e  2 i t

*
*

( b a  a b) 
(
) 
sin 2 t
2
2
2i
2
P460 - Spin
12
Arbitrary Angles
•
can look at any direction (p 160 and problem 10-2 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.
rˆ  sin  cos  iˆ  sin  sin  ˆj  cos  kˆ
• Put components into Pauli spin matrices
 
cos
sin  cos   i sin  sin  

S r  
 cos
 sin  cos   i sin  sin 

• and solve for its eigenvalues

S r    | S  I | 0    1
P460 - Spin
13
rˆ  sin  cos  iˆ  sin  sin  ˆj  cos  kˆ
 
cos
S r  
 sin  cos   i sin  sin 
• Go ahead and solve for eigenspinors.
sin  cos   i sin  sin  

 cos


S r   
a
for   1    
b

 
a cos   b sin  (cos   i sin  )  a
a (1  cos  )
sin  2 i
cos
b
a
e (use 1sin
 tan

i
sin  e
cos  2
• Gives (phi phase is arbitrary)
• if r in z,x,y
directions

2
)
 e  i sin 2 
 cos 2 
r

 for   1     
   i
 



 e sin 2 
  cos 2 
r

-
1
 0 
z :   0    
,





 0
  1

 


x : 

y : 

2
2
,  0   




, 





2
 
1
2
1
2
1
2
i
2
P460 - Spin

 12
,    

 1
2






 i2

,    

 1
2






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:
S
i
, S j   i ijk S k
• representations of angular momentum have 2 quantum numbers:
3
l  0, 1
2 ,1, 2 ......
m  l ,l  1...l  1, l
2l  1 states
• 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

   
 
J  L  S L   Li S   Si



1
if L  0, S  2  J  12 , J z   12
sin glet  doublet  doublet
P460 - Spin
1 2  2
15
Combining Angular Momentum
• Non-trivial examples. add 2 spins. The z-components add “linearly”
and the total adds “vectorally”. Really means add up z-component
and then divide up states into SU(2) groups

if S1 
Jz 
1
2

1
2

, S2 
1
2
1
1
2

1
2
0

1
2

1
2
0

1
2

1
2
 1
1
2
with S z1  
1
2
Sz 2  
1
2
4 terms. need to split
up. The two 0 mix



J  S1  S2



 J  1, J z  1,0,1 AND J  0, J z  0
doublet  doublet  triplet  sin glet
2  2  31
P460 - Spin
16
Combining Angular Momentum
• add spin and orbital angular momentum
if S 
1
2
, L  1 with S z  
Jz  1

3
2
,
1
2

 J 
,
0
1
2
1
2
3
2
,
1
2
,
1
1
2
1
2

, Jz  
,
3
2
1
2
Lz  1,0,1
1
2
3
2
,
1
2

AND J 
1
2
, Jz  
1
2
triplet  doublet  quartet  doublet
3 2  4  2
P460 - Spin
17
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
l
m  total   m
m
  m
m
individual terms.
1
l
m
 total    m1
2
m2
1
   m1
2
m2
m  m1  m2  m1  m2
m1
m2
  l
m1
m2
  l
P460 - Spin
m   l
m    l
m
m
18
Combining Angular Momentum
• Clebsch-Gordon coefficients
• these give the “dot product” or rotation between the total and the
individual terms. “easy” but need to remember what different
quantum number labels refer to
l
l
m
m
 total   m1
m2
 total    m1
m2
  m1
   m1
m2
m2
m  m1  m2  m1  m2
m1
m2
  l
m1
m2
  l
m   l
m    l
P460 - Spin
m
m
19
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
 ,  ,  , 
Sz  
 1  
Sz  
0 
Sz  
 1  
Sz  
S z  S1z  S2 z
0 
S   S1  S2
C   (l  m )( l  m  1)
• step down from ++ state
S1    C ( 12 , 12 )      
S2    C ( 12 , 12 )      
 S   
2 (
  
2
)
S  l  1, m  1  C (1,1) l  1, m  0 
2 l  1, m  0
1
(    )
2
1
 l  0, m  0 
(      ) orthogonal
2
 l  1, m  0 
•
Clebsch-Gordon coefficients 
1
2
P460 - Spin
20
Combining Ang. Momentum
•
•
•
check that eigenstates have right eigenvalue for S2
first write down
2

 2
2
2


S  ( S1  S 2 )  S1  S 2  2 S1  S 2
2
2
 S1  S 2  2 S1 z S 2 z  2 S1 x S 2 x  2 S1 y S 2 y
2
2
 S1  S 2  2 S1 z S 2 z  S1 S 2   S1 S 2 
and then look at terms
S   S x  iS y

S12 X  


1
3 2
(( S12  )   ( S12  )  ) 
 X
4
2
1

3 2
X 
(    )
S 22 X  
 X
2
4


2 S1 z S 2 z X   2( )(  ) X 
2
2
with S 2     S1    S 2     S1    0
and S1 S 2   
 2  
S 2  S1  2  
0
 ( S 2  S1  S1 S 2  ) X     2 X 
•
putting it all together see eigenstates
2
3
S X   2 ( 4

3
4

1
2
 2
 1) X    2 
 0
X
 
P460 - Spin
21
• L=1 + S=1/2
2 terms
Lz
j
m 
3
2
1
2

j
m 
1
2
1
2

1
3
1
sqrt( 2 )
3

J2
Lz
Sz
1

1
2

3
2
3
2
0

1
2

1
2
3
2
1
2
1

1
2

1
2
3
2
1
2
1

1
2

1
2
3
2
1
2
0

1
2

1
2
3
2
1
2
1

1
2

3
2
3
2
• Example of how states “add”:
Jz

J1
Sz

1
1
2


1
2
sqrt( 2 )
3

1
3
0
1
2
0
1
2
• Note Clebsch-Gordon coefficients (used in PHYS 374 class for Mossbauer
spectroscopy).
1 sqrt ( 2)

3
,
3
P460 - Spin
22
• Clebsch-Gordon coefficients for different J,L,S
P460 - Spin
23