Transcript Spin and Magnetic Mon\ments

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

 r 2
but L  m vr


b
q
 2m L  l   gl  L
qv
2r
e
b 
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
L2   l (l  1) 
 l  g l  b l (l  1)
 zl   g l  b ml
P460 - Spin
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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.
KG : E 2  p 2  m 2 D : E 
p 2  m2
• Solve by substituting operators for E,p. The Dirac
equation ends up with magnetic moment terms and
an extra degree of freedom (the spin)
P460 - Spin
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Spin 1/2 expectation values
• similar eigenvalues as orbital angular momentum
(but SU(2)). No 3D “function”
S , S   
i
j
2
2

S

S

s
(
s

1
)

, S z   | s | ... | s |
ijk
k
for S  12  S 2  12 23  2  43  2 , S z   12 , 12 


g s b S
s   
g s  2.00232 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
 1 0


 0 1
1 0 
 0 1
0  i



S z  2 
 S x  2 
 S y  2 

0

1
1
0
i
0







Si   i  S 2 
2
3 2
4
• with two eigenstates (eigenspinors)
1
        S z eigenvalue 2
 0
 0
        S z eigenvalue 2
1
P460 - Spin
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Spin 1/2 expectation values
• “total” spin direction not aligned with any
component.
• can get angle of spin with a component
1

Sz
1
2
cos    

3
3
S

4
P460 - Spin
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Spin 1/2 expectation values
• Let’s assume state in an arbitrary combination of
spin-up and spin-down states.
 a
  a   b     with | a |2  | b |2  1
b
• expectation values. z-component
S z   | S z |   a
*
b

* 
2
 1 0  a 

 
 0  1  b 
 2 ( a 2  b2 )
• x-component
S x  a*

0

2  a 
*
   2 (a*b  b*a )
b  
 2 0  b 
• y-component
S y  a*

0


2 i  a 
*
   i 2 (a*b  b*a )
b  
0  b 
2i
P460 - Spin
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Spin 1/2 expectation values
example

• assume wavefunction is
1
6
• expectation values. z-component
S z  2  a 2 
S z   2  b2 
2
6
1  i 


 2 
4
6
S z  2 (a 2  b2 )   3
• x-component

0

2  a 
t
*
*
 
S x   S x   a b  
 2 0  b 
 2 (a *b  b*a)  2 16 (1  i )2  2(1  i ) 



3
• Can also ask what is the probability to have
different components. As normalized, by inspection
probability ( S x  2 ) 
5
6
probability ( S x   2 )  16  ( 56  16 ) 12  13
• or could rotate wavefunction to basis where x is
diagonal
P460 - Spin
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•
Can also determine

2
4
a
*
S
2
y

2
4
a
*
S
2
z

2
4
a
*
S
2
x
0
b 
1
* 0
b 
i
* 1
b 
0
1  0 1  a   2 *
*
2

   4 ( a a  b b)  4
0  1 0  b 
 i   0  i  a   2 *
*
2

   4 ( a a  b b)  4
0   i 0  b 
0  1 0  a   2 *
*
2

   4 ( a a  b b)  4
 1  0  1   b 
*
S x2  S y2  S z2 
1
3
S2
• and widths
2

( S x ) 2  S x2  S x  (1  ( a *b  b*a ) 2 )
4
2
2

( S y ) 2  S y2  S y  (1  ( a *b  b*a ) 2 )
4
2

2
( S z ) 2  S z2  S z  (1  ( a *a  b*b) 2 )
4
2
P460 - Spin
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Widths- example
• Can look at the widths of spin terms if in a given
 1
eigenstate
    
 0
• z picked as diagonal and so
 1 0  1 0  1  
2


   4
S z  4 1 0
 0  1 0  1 0 
2
(S z )  S
2
2
2
z
 Sz
2

2
4
(1  1)  0
• for off-diagonal
 0 2  1 
   0
S x  1 0 
 2 0  0 
 0 1  0 1  1 
2
2

  
S x  4 1 0
 1 0  1 0  0 
(S x ) 2  S x2  S x
2

P460 - Spin
2
4
2
4
8
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 z-component (also true for l>0)
Sz 
z

2
S2 
3
2
 cos 
1
3
B
S
• Example: external magnetic field. Added energy

E  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 field) with Larmor
g s b B
frequency of


P460 - Spin
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Precession - details
• Hamiltonian for an electron in a magnetic field
  
eg   
H
 B
S 
4m
2
• assume solution of form
   (t ) 

  
   (t ) 
H  i
d
dt
• If B direction defines z-axis have Scr.eq.

1 0 
  B  B

0

1



d     egB  1 0    
i   

 
dt     4m  0  1   
• And can get eigenvalues and eigenfunctions
egB  1 
egB  0 
 
     
  
4m
4m
 0
1
 aeit 
a
 (0)     (t )   it 
b
 be 
P460 - Spin
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Precession - details
• Assume at t=0 in the + eigenstate of Sx
  0 1  a    a   a  1 1

        
 
1
0
b
b
b
2
2 1
  2    
i t
1 e 
 i t
  (t ) 
2e 



• Solve for the x and y expectation values. See how
they precess around the z-axis
 *
 e2it  e 2it

*
S x  (a b  b a )  (
)  cos2 t
2
2
2
2
i *
 e2it  e 2it

*
S y  ( b a  a b)  (
)  sin 2 t
2
2
2i
2
P460 - Spin
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Arbitrary Angles
•
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.
rˆ  sin cos iˆ  sin sin  ˆj  cos kˆ
• Put components into Pauli spin matrices
 
cos
sin  cos  i sin  sin  

Sr  
 cos
 sin  cos  i sin  sin 

• and solve for its eigenvalues

Sr    | S  I | 0    1
P460 - Spin
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rˆ  sin cos iˆ  sin sin  ˆj  cos kˆ
 
cos
sin  cos  i sin  sin  

Sr  
 cos
 sin  cos  i sin  sin 

• Go ahead and solve for eigenspinors.

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
 t an 2 )

i
sin  e
cos 2
• Phi phase is arbitrary. gives
 i

 

cos


e
sin
2

 r   i 2   for   1   r  
 
 e sin 2 
  cos 2 
• if r in z,x,y-directions  1 
 0
z :   0      ,     
 0
  1


x :   2 ,   0     

1
2
1
2

 12 
,     1 

 

 2

    

P460 - Spin
1
2
i
2
 i2 

,     1 

 

 2 13
y :   2 ,   2
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 , S   i
i
j
S
ijk k
• representations of angular momentum have 2
quantum numbers:
l  0, 12 ,1, 32 ......
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 1  2  2
P460 - Spin
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Combining Angular Momentum
• Non-trivial examples. add 2 spins


1
if S1  2 , S2  12 with S z1   12 S z 2   12
  
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
• add spin and orbital angular momentum


if L  1, S  12 with Lz  1,0,1 S z   12



3
3
1
 J  2 , J z   2 , 2 AND J  12 , J z   12
triplet  doublet quartet doublet
3 2  4 2
P460 - Spin
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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.
l
m  total   m1 m2   m1 m2
l  m  total    m1 m2    m1 m2
m  m1  m2  m1  m2
m1 m2   l
m   l m
m1 m2    l m    l  m
P460 - Spin
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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
 ,  ,  , 
S z  S1z  S2 z
S z    1  
Sz    0  
S  S1  S2
Sz    0  
C   (l  m)(l  m  1)
S z    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
P460 - Spin

1
2
17
Combining Ang. Momentum
• check that eigenstates have right eigenvalue for S2
• first write down
S  Sx  iSy
2
  2 2 2
 
S  ( S1  S2 )  S1  S2  2 S1  S2
2 2
 S1  S2  2 S1z S2 z  2 S1x S2 x  2 S1 y S2 y
2 2
 S1  S2  2 S1z S2 z  S1 S2  S1 S2
• and then look at terms
2
2
1 2
3
S1 X  
((S1  )   ( S1  )  )   2 X 
4
2
1

3
X 
(    )
S22 X    2 X 
2
4


2 S1z S2 z X   2( )( ) X 
2
2
with S2    S1    S2    S1    0
and S1 S2     2   S2 S1  2    0
 ( S2 S1  S1 S2 ) X     2 X 
• putting it all together see eigenstates
2
2 3
2  2
3
1
S X    ( 4  4  2  1) X      X 
 0
P460 - Spin
18
• L=1 + S=1/2
Jz

J1
 1  12  32
3
2
 12  12
3
2
1
2
 1  12  12
3
2
1
2
 1  12  12
3
2
1
2
 12  12
3
2
1
2
 1  12  32
3
2
Lz
Sz
0
2 terms
0

J2
• Example of how states “add”:
Lz
Sz
j m 
3
2
1
2

1
3
1  12 
2
3
0
1
2
j m 
1
2
1
2

2
3
1  12 
1
3
0
1
2
• Note Clebsch-Gordon coefficients
1 2

,
3 3
P460 - Spin
19
• Clebsch-Gordon coefficients for different J,L,S
P460 - Spin
20